Contents

Diagnostic messages

Contents

This page lists diagnostic messages produced by the Dart analyzer, with details about what those messages mean and how you can fix your code. For more information about the analyzer, see Customizing static analysis.

Glossary

This page uses the following terms:

Constant context

A constant context is a region of code in which it isn’t necessary to include the const keyword because it’s implied by the fact that everything in that region is required to be a constant. The following locations are constant contexts:

  • Everything inside a list, map or set literal that’s prefixed by the const keyword. Example:

    var l = const [/*constant context*/];
    
  • The arguments inside an invocation of a constant constructor. Example:

    var p = const Point(/*constant context*/);
    
  • The initializer for a variable that’s prefixed by the const keyword. Example:

    const v = /*constant context*/;
    
  • Annotations

  • The expression in a case clause. Example:

    void f(int e) {
      switch (e) {
        case /*constant context*/:
          break;
      }
    }
    

Definite assignment

Definite assignment analysis is the process of determining, for each local variable at each point in the code, which of the following is true:

  • The variable has definitely been assigned a value (definitely assigned).
  • The variable has definitely not been assigned a value (definitely unassigned).
  • The variable might or might not have been assigned a value, depending on the execution path taken to arrive at that point.

Definite assignment analysis helps find problems in code, such as places where a variable that might not have been assigned a value is being referenced, or places where a variable that can only be assigned a value one time is being assigned after it might already have been assigned a value.

For example, in the following code the variable s is definitely unassigned when it’s passed as an argument to print:

void f() {
  late String s;
  print(s);
}

But in the following code, the variable s is definitely assigned:

void f(String name) {
  String s = 'Hello $name!';
  print(s);
}

Definite assignment analysis can even tell whether a variable is definitely assigned (or unassigned) when there are multiple possible execution paths. In the following code the print function is called if execution goes through either the true or the false branch of the if statement, but because s is assigned no matter which branch is taken, it’s definitely assigned before it’s passed to print:

void f(String name, bool casual) {
  late String s;
  if (casual) {
    s = 'Hi $name!';
  } else {
    s = 'Hello $name!';
  }
  print(s);
}

In flow analysis, the end of the if statement is referred to as a join—a place where two or more execution paths merge back together. Where there’s a join, the analysis says that a variable is definitely assigned if it’s definitely assigned along all of the paths that are merging, and definitely unassigned if it’s definitely unassigned along all of the paths.

Sometimes a variable is assigned a value on one path but not on another, in which case the variable might or might not have been assigned a value. In the following example, the true branch of the if statement might or might not be executed, so the variable might or might be assigned a value:

void f(String name, bool casual) {
  late String s;
  if (casual) {
    s = 'Hi $name!';
  }
  print(s);
}

The same is true if there is a false branch that doesn’t assign a value to s.

The analysis of loops is a little more complicated, but it follows the same basic reasoning. For example, the condition in a while loop is always executed, but the body might or might not be. So just like an if statement, there’s a join at the end of the while statement between the path in which the condition is true and the path in which the condition is false.

For additional details, see the [specification of definite assignment][definiteAssignmentSpec].

definiteAssignmentSpec

Override inference

Override inference is the process by which any missing types in a method declaration are inferred based on the corresponding types from the method or methods that it overrides.

If a candidate method (the method that’s missing type information) overrides a single inherited method, then the corresponding types from the overridden method are inferred. For example, consider the following code:

class A {
  int m(String s) => 0;
}

class B extends A {
  @override
  m(s) => 1;
}

The declaration of m in B is a candidate because it’s missing both the return type and the parameter type. Because it overrides a single method (the method m in A), the types from the overridden method will be used to infer the missing types and it will be as if the method in B had been declared as int m(String s) => 1;.

If a candidate method overrides multiple methods, and the function type one of those overridden methods, Ms, is a supertype of the function types of all of the other overridden methods, then Ms is used to infer the missing types. For example, consider the following code:

class A {
  int m(num n) => 0;
}

class B {
  num m(int i) => 0;
}

class C implements A, B {
  @override
  m(n) => 1;
}

The declaration of m in C is a candidate for override inference because it’s missing both the return type and the parameter type. It overrides both m in A and m in B, so we need to choose one of them from which the missing types can be inferred. But because the function type of m in A (int Function(num)) is a supertype of the function type of m in B (num Function(int)), the function in A is used to infer the missing types. The result is the same as declaring the method in C as int m(num n) => 1;.

It is an error if none of the overridden methods has a function type that is a supertype of all the other overridden methods.

Potentially non-nullable

A type is potentially non-nullable if it’s either explicitly non-nullable or if it’s a type parameter.

A type is explicitly non-nullable if it is a type name that isn’t followed by a question mark. Note that there are a few types that are always nullable, such as Null and dynamic, and that FutureOr is only non-nullable if it isn’t followed by a question mark and the type argument is non-nullable (such as FutureOr<String>).

Type parameters are potentially non-nullable because the actual runtime type (the type specified as a type argument) might be non-nullable. For example, given a declaration of class C<T> {}, the type C could be used with a non-nullable type argument as in C<int>.

Diagnostics

The analyzer produces the following diagnostics for code that doesn’t conform to the language specification or that might work in unexpected ways.

abstract_super_member_reference

The {0} ‘{1}’ is always abstract in the supertype.

Description

The analyzer produces this diagnostic when an inherited member is referenced using super, but there is no concrete implementation of the member in the superclass chain. Abstract members can’t be invoked.

Examples

The following code produces this diagnostic because B doesn’t inherit a concrete implementation of a:

abstract class A {
  int get a;
}
class B extends A {
  int get a => super.a;
}

Common fixes

Remove the invocation of the abstract member, possibly replacing it with an invocation of a concrete member.

ambiguous_extension_member_access

A member named ‘{0}’ is defined in extensions ‘{1}’ and ‘{2}’ and neither is more specific.

Description

When code refers to a member of an object (for example, o.m() or o.m or o[i]) where the static type of o doesn’t declare the member (m or [], for example), then the analyzer tries to find the member in an extension. For example, if the member is m, then the analyzer looks for extensions that declare a member named m and have an extended type that the static type of o can be assigned to. When there’s more than one such extension in scope, the extension whose extended type is most specific is selected.

The analyzer produces this diagnostic when none of the extensions has an extended type that’s more specific than the extended types of all of the other extensions, making the reference to the member ambiguous.

Examples

The following code produces this diagnostic because there’s no way to choose between the member in E1 and the member in E2:

extension E1 on String {
  int get charCount => 1;
}

extension E2 on String {
  int get charCount => 2;
}

void f(String s) {
  print(s.charCount);
}

Common fixes

If you don’t need both extensions, then you can delete or hide one of them.

If you need both, then explicitly select the one you want to use by using an extension override:

extension E1 on String {
  int get charCount => length;
}

extension E2 on String {
  int get charCount => length;
}

void f(String s) {
  print(E2(s).charCount);
}

ambiguous_import

The name ‘{0}’ is defined in the libraries {1}.

Description

The analyzer produces this diagnostic when a name is referenced that is declared in two or more imported libraries.

Examples

Given a library (a.dart) that defines a class (C in this example):

class A {}
class C {}

And a library (b.dart) that defines a different class with the same name:

class B {}
class C {}

The following code produces this diagnostic:

import 'a.dart';
import 'b.dart';

void f(C c1, C c2) {}

Common fixes

If any of the libraries aren’t needed, then remove the import directives for them:

import 'a.dart';

void f(C c1, C c2) {}

If the name is still defined by more than one library, then add a hide clause to the import directives for all except one library:

import 'a.dart' hide C;
import 'b.dart';

void f(C c1, C c2) {}

If you must be able to reference more than one of these types, then add a prefix to each of the import directives, and qualify the references with the appropriate prefix:

import 'a.dart' as a;
import 'b.dart' as b;

void f(a.C c1, b.C c2) {}

ambiguous_set_or_map_literal_both

This literal contains both ‘Map’ and ‘Iterable’ spreads, which makes it impossible to determine whether the literal is a map or a set.

Description

Because map and set literals use the same delimiters ({ and }), the analyzer looks at the type arguments and the elements to determine which kind of literal you meant. When there are no type arguments and all of the elements are spread elements (which are allowed in both kinds of literals), then the analyzer uses the types of the expressions that are being spread. If all of the expressions have the type Iterable, then it’s a set literal; if they all have the type Map, then it’s a map literal.

The analyzer produces this diagnostic when some of the expressions being spread have the type Iterable and others have the type Map, making it impossible for the analyzer to determine whether you are writing a map literal or a set literal.

Examples

The following code produces this diagnostic:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, ...b, ...c};

The list b can only be spread into a set, and the maps a and c can only be spread into a map, and the literal can’t be both.

Common fixes

There are two common ways to fix this problem. The first is to remove all of the spread elements of one kind or another, so that the elements are consistent. In this case, that likely means removing the list and deciding what to do about the now unused parameter:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, ...c};

The second fix is to change the elements of one kind into elements that are consistent with the other elements. For example, you can add the elements of the list as keys that map to themselves:

union(Map<String, String> a, List<String> b, Map<String, String> c) =>
    {...a, for (String s in b) s: s, ...c};

ambiguous_set_or_map_literal_either

This literal must be either a map or a set, but the elements don’t have enough information for type inference to work.

Description

Because map and set literals use the same delimiters ({ and }), the analyzer looks at the type arguments and the elements to determine which kind of literal you meant. When there are no type arguments and all of the elements are spread elements (which are allowed in both kinds of literals) then the analyzer uses the types of the expressions that are being spread. If all of the expressions have the type Iterable, then it’s a set literal; if they all have the type Map, then it’s a map literal.

This diagnostic is produced when none of the expressions being spread have a type that allows the analyzer to decide whether you were writing a map literal or a set literal.

Examples

The following code produces this diagnostic:

union(a, b) => {...a, ...b};

The problem occurs because there are no type arguments, and there is no information about the type of either a or b.

Common fixes

There are three common ways to fix this problem. The first is to add type arguments to the literal. For example, if the literal is intended to be a map literal, you might write something like this:

union(a, b) => <String, String>{...a, ...b};

The second fix is to add type information so that the expressions have either the type Iterable or the type Map. You can add an explicit cast or, in this case, add types to the declarations of the two parameters:

union(List<int> a, List<int> b) => {...a, ...b};

The third fix is to add context information. In this case, that means adding a return type to the function:

Set<String> union(a, b) => {...a, ...b};

In other cases, you might add a type somewhere else. For example, say the original code looks like this:

union(a, b) {
  var x = {...a, ...b};
  return x;
}

You might add a type annotation on x, like this:

union(a, b) {
  Map<String, String> x = {...a, ...b};
  return x;
}

argument_type_not_assignable

The argument type ‘{0}’ can’t be assigned to the parameter type ‘{1}’.

Description

The analyzer produces this diagnostic when the static type of an argument can’t be assigned to the static type of the corresponding parameter.

Examples

The following code produces this diagnostic because a num can’t be assigned to a String:

String f(String x) => x;
String g(num y) => f(y);

Common fixes

If possible, rewrite the code so that the static type is assignable. In the example above you might be able to change the type of the parameter y:

String f(String x) => x;
String g(String y) => f(y);

If that fix isn’t possible, then add code to handle the case where the argument value isn’t the required type. One approach is to coerce other types to the required type:

String f(String x) => x;
String g(num y) => f(y.toString());

Another approach is to add explicit type tests and fallback code:

String f(String x) => x;
String g(num y) => f(y is String ? y : '');

If you believe that the runtime type of the argument will always be the same as the static type of the parameter, and you’re willing to risk having an exception thrown at runtime if you’re wrong, then add an explicit cast:

String f(String x) => x;
String g(num y) => f(y as String);

assignment_to_final

‘{0}’ can’t be used as a setter because it’s final.

Description

The analyzer produces this diagnostic when it finds an invocation of a setter, but there’s no setter because the field with the same name was declared to be final or const.

Examples

The following code produces this diagnostic because v is final:

class C {
  final v = 0;
}

f(C c) {
  c.v = 1;
}

Common fixes

If you need to be able to set the value of the field, then remove the modifier final from the field:

class C {
  int v = 0;
}

f(C c) {
  c.v = 1;
}

assignment_to_final_local

The final variable ‘{0}’ can only be set once.

Description

The analyzer produces this diagnostic when a local variable that was declared to be final is assigned after it was initialized.

Examples

The following code produces this diagnostic because x is final, so it can’t have a value assigned to it after it was initialized:

void f() {
  final x = 0;
  x = 3;
  print(x);
}

Common fixes

Remove the keyword final, and replace it with var if there’s no type annotation:

void f() {
  var x = 0;
  x = 3;
  print(x);
}

assignment_to_final_no_setter

There isn’t a setter named ‘{0}’ in class ‘{1}’.

Description

The analyzer produces this diagnostic when a reference to a setter is found; there is no setter defined for the type; but there is a getter defined with the same name.

Examples

The following code produces this diagnostic because there is no setter named x in C, but there is a getter named x:

class C {
  int get x => 0;
  set y(int p) {}
}

void f(C c) {
  c.x = 1;
}

Common fixes

If you want to invoke an existing setter, then correct the name:

class C {
  int get x => 0;
  set y(int p) {}
}

void f(C c) {
  c.y = 1;
}

If you want to invoke the setter but it just doesn’t exist yet, then declare it:

class C {
  int get x => 0;
  set x(int p) {}
  set y(int p) {}
}

void f(C c) {
  c.x = 1;
}

assignment_to_method

Methods can’t be assigned a value.

Description

The analyzer produces this diagnostic when the target of an assignment is a method.

Examples

The following code produces this diagnostic because f can’t be assigned a value because it’s a method:

class C {
  void f() {}

  void g() {
    f = null;
  }
}

Common fixes

Rewrite the code so that there isn’t an assignment to a method.

await_in_late_local_variable_initializer

The ‘await’ expression can’t be used in a ‘late’ local variable’s initializer.

Description

The analyzer produces this diagnostic when a local variable that has the ‘late’ modifier uses an ‘await’ expression in the initializer.

Example

The following code produces this diagnostic because an ‘await’ expression is used in the initializer for ‘v’, a local variable that is marked ‘late’:

Future<int> f() async {
  late v = await 42;
  return v;
}

Common fixes

If the initializer can be rewritten to not use ‘await’, then rewrite it:

Future<int> f() async {
  late v = 42;
  return v;
}

If the initializer can’t be rewritten, then remove the ‘late’ modifier:

Future<int> f() async {
  var v = await 42;
  return v;
}

body_might_complete_normally

The body might complete normally, causing ‘null’ to be returned, but the return type is a potentially non-nullable type.

Description

The analyzer produces this diagnostic when a method or function has a return type that’s potentially non-nullable but would implicitly return null if control reached the end of the function.

Example

The following code produces this diagnostic because the method m has an implicit return of null inserted at the end of the method, but the method is declared to not return null:

class C {
  int m(int t) {
    print(t);
  }
}

The following code produces this diagnostic because the method m has an implicit return of null inserted at the end of the method, but because the class C can be instantiated with a non-nullable type argument, the method is effectively declared to not return null:

class C<T> {
  T m(T t) {
    print(t);
  }
}

Common fixes

If there’s a reasonable value that can be returned, then add a return statement at the end of the method:

class C<T> {
  T m(T t) {
    print(t);
    return t;
  }
}

If the method won’t reach the implicit return, then add a throw at the end of the method:

class C<T> {
  T m(T t) {
    print(t);
    throw '';
  }
}

If the method intentionally returns null at the end, then change the return type so that it’s valid to return null:

class C<T> {
  T? m(T t) {
    print(t);
  }
}

built_in_identifier_as_extension_name

The built-in identifier ‘{0}’ can’t be used as an extension name.

Description

The analyzer produces this diagnostic when the name of an extension is a built-in identifier. Built-in identifiers can’t be used as extension names.

Examples

The following code produces this diagnostic because mixin is a built-in identifier:

extension mixin on int {}

Common fixes

Choose a different name for the extension.

built_in_identifier_as_type

The built-in identifier ‘{0}’ can’t be used as a type.

Description

The analyzer produces this diagnostic when a built-in identifier is used where a type name is expected.

Examples

The following code produces this diagnostic because import can’t be used as a type because it’s a built-in identifier:

import<int> x;

Common fixes

Replace the built-in identifier with the name of a valid type:

List<int> x;

case_block_not_terminated

The last statement of the ‘case’ should be ‘break’, ‘continue’, ‘rethrow’, ‘return’, or ‘throw’.

Description

The analyzer produces this diagnostic when the last statement in a case block isn’t one of the required terminators: break, continue, rethrow, return, or throw.

Examples

The following code produces this diagnostic because the case block ends with an assignment:

void f(int x) {
  switch (x) {
    case 0:
      x += 2;
    default:
      x += 1;
  }
}

Common fixes

Add one of the required terminators:

void f(int x) {
  switch (x) {
    case 0:
      x += 2;
      break;
    default:
      x += 1;
  }
}

case_expression_type_is_not_switch_expression_subtype

The switch case expression type ‘{0}’ must be a subtype of the switch expression type ‘{1}’.

Description

The analyzer produces this diagnostic when the expression following case in a switch statement has a static type that isn’t a subtype of the static type of the expression following switch.

Example

The following code produces this diagnostic because 1 is an int, which isn’t a subtype of String (the type of s):

void f(String s) {
  switch (s) {
    case 1:
      break;
  }
}

Common fixes

If the value of the case expression is wrong, then change the case expression so that it has the required type:

void f(String s) {
  switch (s) {
    case '1':
      break;
  }
}

If the value of the case expression is correct, then change the switch expression to have the required type:

void f(int s) {
  switch (s) {
    case 1:
      break;
  }
}

cast_to_non_type

The name ‘{0}’ isn’t a type, so it can’t be used in an ‘as’ expression.

Description

The analyzer produces this diagnostic when the name following the as in a cast expression is defined to be something other than a type.

Examples

The following code produces this diagnostic because x is a variable, not a type:

num x = 0;
int y = x as x;

Common fixes

Replace the name with the name of a type:

num x = 0;
int y = x as int;

concrete_class_with_abstract_member

‘{0}’ must have a method body because ‘{1}’ isn’t abstract.

Description

The analyzer produces this diagnostic when a member of a concrete class is found that doesn’t have a concrete implementation. Concrete classes aren’t allowed to contain abstract members.

Examples

The following code produces this diagnostic because m is an abstract method but C isn’t an abstract class:

class C {
  void m();
}

Common fixes

If it’s valid to create instances of the class, provide an implementation for the member:

class C {
  void m() {}
}

If it isn’t valid to create instances of the class, mark the class as being abstract:

abstract class C {
  void m();
}

const_constructor_param_type_mismatch

A value of type ‘{0}’ can’t be assigned to a parameter of type ‘{1}’ in a const constructor.

Description

The analyzer produces this diagnostic when the runtime type of a constant value can’t be assigned to the static type of a constant constructor’s parameter.

Example

The following code produces this diagnostic because the runtime type of i is int, which can’t be assigned to the static type of s:

class C {
  final String s;

  const C(this.s);
}

const dynamic i = 0;

void f() {
  const C(i);
}

Common fixes

Pass a value of the correct type to the constructor:

class C {
  final String s;

  const C(this.s);
}

const dynamic i = 0;

void f() {
  const C('$i');
}

const_constructor_with_field_initialized_by_non_const

Can’t define the ‘const’ constructor because the field ‘{0}’ is initialized with a non-constant value.

Description

The analyzer produces this diagnostic when a constructor has the keyword const, but a field in the class is initialized to a non-constant value.

Example

The following code produces this diagnostic because the field s is initialized to a non-constant value:

class C {
  final String s = 3.toString();
  const C();
}

Common fixes

If the field can be initialized to a constant value, then change the initializer to a constant expression:

class C {
  final String s = '3';
  const C();
}

If the field can’t be initialized to a constant value, then remove the keyword const from the constructor:

class C {
  final String s = 3.toString();
  C();
}

const_constructor_with_non_final_field

Can’t define a const constructor for a class with non-final fields.

Description

The analyzer produces this diagnostic when a constructor is marked as a const constructor, but the constructor is defined in a class that has at least one non-final instance field (either directly or by inheritance).

Examples

The following code produces this diagnostic because the field x isn’t final:

class C {
  int x;

  const C(this.x);
}

Common fixes

If it’s possible to mark all of the fields as final, then do so:

class C {
  final int x;

  const C(this.x);
}

If it isn’t possible to mark all of the fields as final, then remove the keyword const from the constructor:

class C {
  int x;

  C(this.x);
}

const_initialized_with_non_constant_value

Const variables must be initialized with a constant value.

Description

The analyzer produces this diagnostic when a value that isn’t statically known to be a constant is assigned to a variable that’s declared to be a ‘const’ variable.

Examples

The following code produces this diagnostic because x isn’t declared to be const:

var x = 0;
const y = x;

Common fixes

If the value being assigned can be declared to be const, then change the declaration:

const x = 0;
const y = x;

If the value can’t be declared to be const, then remove the const modifier from the variable, possibly using final in its place:

var x = 0;
final y = x;

const_instance_field

Only static fields can be declared as const.

Description

The analyzer produces this diagnostic when an instance field is marked as being const.

Examples

The following code produces this diagnostic because f is an instance field:

class C {
  const int f = 3;
}

Common fixes

If the field needs to be an instance field, then remove the keyword const, or replace it with final:

class C {
  final int f = 3;
}

If the field really should be a const field, then make it a static field:

class C {
  static const int f = 3;
}

const_not_initialized

The constant ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a variable that is declared to be a constant doesn’t have an initializer.

Examples

The following code produces this diagnostic because c isn’t initialized:

const c;

Common fixes

Add an initializer:

const c = 'c';

const_spread_expected_list_or_set

A list or a set is expected in this spread.

Description

The analyzer produces this diagnostic when the expression of a spread operator in a constant list or set evaluates to something other than a list or a set.

Examples

The following code produces this diagnostic because the value of list1 is null, which is neither a list nor a set:

const List<int> list1 = null;
const List<int> list2 = [...list1];

Common fixes

Change the expression to something that evaluates to either a constant list or a constant set:

const List<int> list1 = [];
const List<int> list2 = [...list1];

const_spread_expected_map

A map is expected in this spread.

Description

The analyzer produces this diagnostic when the expression of a spread operator in a constant map evaluates to something other than a map.

Examples

The following code produces this diagnostic because the value of map1 is null, which isn’t a map:

const Map<String, int> map1 = null;
const Map<String, int> map2 = {...map1};

Common fixes

Change the expression to something that evaluates to a constant map:

const Map<String, int> map1 = {};
const Map<String, int> map2 = {...map1};

const_with_non_const

The constructor being called isn’t a const constructor.

Description

The analyzer produces this diagnostic when the keyword const is used to invoke a constructor that isn’t marked with const.

Examples

The following code produces this diagnostic because the constructor in A isn’t a const constructor:

class A {
  A();
}

A f() => const A();

Common fixes

If it’s desirable and possible to make the class a constant class (by making all of the fields of the class, including inherited fields, final), then add the keyword const to the constructor:

class A {
  const A();
}

A f() => const A();

Otherwise, remove the keyword const:

class A {
  A();
}

A f() => A();

const_with_non_constant_argument

Arguments of a constant creation must be constant expressions.

Description

The analyzer produces this diagnostic when a const constructor is invoked with an argument that isn’t a constant expression.

Examples

The following code produces this diagnostic because i isn’t a constant:

class C {
  final int i;
  const C(this.i);
}
C f(int i) => const C(i);

Common fixes

Either make all of the arguments constant expressions, or remove the const keyword to use the non-constant form of the constructor:

class C {
  final int i;
  const C(this.i);
}
C f(int i) => C(i);

creation_with_non_type

The name ‘{0}’ isn’t a class.

The name ‘{0}’ isn’t a class.

Description

The analyzer produces this diagnostic when an instance creation using either new or const specifies a name that isn’t defined as a class.

Example

The following code produces this diagnostic because f is a function rather than a class:

int f() => 0;

void g() {
  new f();
}

Common fixes

If a class should be created, then replace the invalid name with the name of a valid class:

int f() => 0;

void g() {
  new Object();
}

If the name is the name of a function and you want that function to be invoked, then remove the new or const keyword:

int f() => 0;

void g() {
  f();
}

dead_code

Dead code.

Description

The analyzer produces this diagnostic when code is found that won’t be executed because execution will never reach the code.

Examples

The following code produces this diagnostic because the invocation of print occurs after the function has returned:

void f() {
  return;
  print('here');
}

Common fixes

If the code isn’t needed, then remove it:

void f() {
  return;
}

If the code needs to be executed, then either move the code to a place where it will be executed:

void f() {
  print('here');
  return;
}

Or, rewrite the code before it, so that it can be reached:

void f({bool skipPrinting = true}) {
  if (skipPrinting) {
    return;
  }
  print('here');
}

dead_code_catch_following_catch

Dead code: Catch clauses after a ‘catch (e)’ or an ‘on Object catch (e)’ are never reached.

Description

The analyzer produces this diagnostic when a catch clause is found that can’t be executed because it’s after a catch clause of the form catch (e) or on Object catch (e). The first catch clause that matches the thrown object is selected, and both of those forms will match any object, so no catch clauses that follow them will be selected.

Examples

The following code produces this diagnostic:

void f() {
  try {
  } catch (e) {
  } on String {
  }
}

Common fixes

If the clause should be selectable, then move the clause before the general clause:

void f() {
  try {
  } on String {
  } catch (e) {
  }
}

If the clause doesn’t need to be selectable, then remove it:

void f() {
  try {
  } catch (e) {
  }
}

dead_code_on_catch_subtype

Dead code: This on-catch block won’t be executed because ‘{0}’ is a subtype of ‘{1}’ and hence will have been caught already.

Description

The analyzer produces this diagnostic when a catch clause is found that can’t be executed because it is after a catch clause that catches either the same type or a supertype of the clause’s type. The first catch clause that matches the thrown object is selected, and the earlier clause always matches anything matchable by the highlighted clause, so the highlighted clause will never be selected.

Examples

The following code produces this diagnostic:

void f() {
  try {
  } on num {
  } on int {
  }
}

Common fixes

If the clause should be selectable, then move the clause before the general clause:

void f() {
  try {
  } on int {
  } on num {
  }
}

If the clause doesn’t need to be selectable, then remove it:

void f() {
  try {
  } on num {
  }
}

dead_null_aware_expression

The left operand can’t be null, so the right operand is never executed.

Description

The analyzer produces this diagnostic in two cases.

The first is when the left operand of an ?? operator can’t be null. The right operand is only evaluated if the left operand has the value null, and because the left operand can’t be null, the right operand is never evaluated.

The second is when the left-hand side of an assignment using the ??= operator can’t be null. The right-hand side is only evaluated if the left-hand side has the value null, and because the left-hand side can’t be null, the right-hand side is never evaluated.

Example

The following code produces this diagnostic because x can’t be null:

int f(int x) {
  return x ?? 0;
}

The following code produces this diagnostic because f can’t be null:

class C {
  int f = -1;

  void m(int x) {
    f ??= x;
  }
}

Common fixes

If the diagnostic is reported for an ?? operator, then remove the ?? operator and the right operand:

int f(int x) {
  return x;
}

If the diagnostic is reported for an assignment, and the assignment isn’t needed, then remove the assignment:

class C {
  int f = -1;

  void m(int x) {
  }
}

If the assignment is needed, but should be based on a different condition, then rewrite the code to use = and the different condition:

class C {
  int f = -1;

  void m(int x) {
    if (f < 0) {
      f = x;
    }
  }
}

default_list_constructor

The default ‘List’ constructor isn’t available when null safety is enabled.

Description

The analyzer produces this diagnostic when it finds a use of the default constructor for the class List in code that has opted in to null safety.

Example

Assuming the following code is opted in to null safety, it produces this diagnostic because it uses the default List constructor:

var l = List<int>();

Common fixes

If no initial size is provided, then convert the code to use a list literal:

var l = <int>[];

If an initial size needs to be provided and there is a single reasonable initial value for the elements, then use List.filled:

var l = List.filled(3, 0);

If an initial size needs to be provided but each element needs to be computed, then use List.generate:

var l = List.generate(3, (i) => i);

default_value_in_function_type

Parameters in a function type can’t have default values.

Description

The analyzer produces this diagnostic when a function type associated with a parameter includes optional parameters that have a default value. This isn’t allowed because the default values of parameters aren’t part of the function’s type, and therefore including them doesn’t provide any value.

Example

The following code produces this diagnostic because the parameter p has a default value even though it’s part of the type of the parameter g:

void f(void Function([int p = 0]) g) {
}

Common fixes

Remove the default value from the function-type’s parameter:

void f(void Function([int p]) g) {
}

definitely_unassigned_late_local_variable

The late local variable ‘{0}’ is definitely unassigned at this point.

Description

The analyzer produces this diagnostic when definite assignment analysis shows that a local variable that’s marked as late is read before being assigned.

Example

The following code produces this diagnostic because x wasn’t assigned a value before being read:

void f(bool b) {
  late int x;
  print(x);
}

Common fixes

Assign a value to the variable before reading from it:

void f(bool b) {
  late int x;
  x = b ? 1 : 0;
  print(x);
}

deprecated_member_use

‘{0}’ is deprecated and shouldn’t be used.

‘{0}’ is deprecated and shouldn’t be used. {1}.

Description

The analyzer produces this diagnostic when a deprecated library or class member is used in a different package.

Examples

If the method m in the class C is annotated with @deprecated, then the following code produces this diagnostic:

void f(C c) {
  c.m();
}

Common fixes

The documentation for declarations that are annotated with @deprecated should indicate what code to use in place of the deprecated code.

deprecated_member_use_from_same_package

‘{0}’ is deprecated and shouldn’t be used.

‘{0}’ is deprecated and shouldn’t be used. {1}.

Description

The analyzer produces this diagnostic when a deprecated library member or class member is used in the same package in which it’s declared.

Examples

The following code produces this diagnostic because x is deprecated:

@deprecated
var x = 0;
var y = x;

Common fixes

The fix depends on what’s been deprecated and what the replacement is. The documentation for deprecated declarations should indicate what code to use in place of the deprecated code.

duplicate_constructor

The constructor with name ‘{0}’ is already defined.

The default constructor is already defined.

Description

The analyzer produces this diagnostic when a class declares more than one unnamed constructor or when it declares more than one constructor with the same name.

Examples

The following code produces this diagnostic because there are two declarations for the unnamed constructor:

class C {
  C();

  C();
}

The following code produces this diagnostic because there are two declarations for the constructor named m:

class C {
  C.m();

  C.m();
}

Common fixes

If there are multiple unnamed constructors and all of the constructors are needed, then give all of them, or all except one of them, a name:

class C {
  C();

  C.n();
}

If there are multiple unnamed constructors and all except one of them are unneeded, then remove the constructors that aren’t needed:

class C {
  C();
}

If there are multiple named constructors and all of the constructors are needed, then rename all except one of them:

class C {
  C.m();

  C.n();
}

If there are multiple named constructors and all except one of them are unneeded, then remove the constructorsthat aren’t needed:

class C {
  C.m();
}

duplicate_definition

The name ‘{0}’ is already defined.

Description

The analyzer produces this diagnostic when a name is declared, and there is a previous declaration with the same name in the same scope.

Examples

The following code produces this diagnostic because the name x is declared twice:

int x = 0;
int x = 1;

Common fixes

Choose a different name for one of the declarations.

int x = 0;
int y = 1;

duplicate_import

Duplicate import.

Description

The analyzer produces this diagnostic when an import directive is found that is the same as an import before it in the file. The second import doesn’t add value and should be removed.

Examples

The following code produces this diagnostic:

import 'package:meta/meta.dart';
import 'package:meta/meta.dart';

@sealed class C {}

Common fixes

Remove the unnecessary import:

import 'package:meta/meta.dart';

@sealed class C {}

duplicate_named_argument

The argument for the named parameter ‘{0}’ was already specified.

Description

The analyzer produces this diagnostic when an invocation has two or more named arguments that have the same name.

Examples

The following code produces this diagnostic because there are two arguments with the name a:

void f(C c) {
  c.m(a: 0, a: 1);
}

class C {
  void m({int a, int b}) {}
}

Common fixes

If one of the arguments should have a different name, then change the name:

void f(C c) {
  c.m(a: 0, b: 1);
}

class C {
  void m({int a, int b}) {}
}

If one of the arguments is wrong, then remove it:

void f(C c) {
  c.m(a: 1);
}

class C {
  void m({int a, int b}) {}
}

equal_elements_in_const_set

Two elements in a constant set literal can’t be equal.

Description

The analyzer produces this diagnostic when two elements in a constant set literal have the same value. The set can only contain each value once, which means that one of the values is unnecessary.

Examples

The following code produces this diagnostic because the string 'a' is specified twice:

const Set<String> set = {'a', 'a'};

Common fixes

Remove one of the duplicate values:

const Set<String> set = {'a'};

Note that literal sets preserve the order of their elements, so the choice of which element to remove might affect the order in which elements are returned by an iterator.

equal_elements_in_set

Two elements in a set literal shouldn’t be equal.

Description

The analyzer produces this diagnostic when an element in a non-constant set is the same as a previous element in the same set. If two elements are the same, then the second value is ignored, which makes having both elements pointless and likely signals a bug.

Example

The following code produces this diagnostic because the element 1 appears twice:

const a = 1;
const b = 1;
var s = <int>{a, b};

Common fixes

If both elements should be included in the set, then change one of the elements:

const a = 1;
const b = 2;
var s = <int>{a, b};

If only one of the elements is needed, then remove the one that isn’t needed:

const a = 1;
var s = <int>{a};

Note that literal sets preserve the order of their elements, so the choice of which element to remove might affect the order in which elements are returned by an iterator.

equal_keys_in_const_map

Two keys in a constant map literal can’t be equal.

Description

The analyzer produces this diagnostic when a key in a constant map is the same as a previous key in the same map. If two keys are the same, then the second value would overwrite the first value, which makes having both pairs pointless.

Examples

The following code produces this diagnostic because the key 1 is used twice:

const map = <int, String>{1: 'a', 2: 'b', 1: 'c', 4: 'd'};

Common fixes

If both entries should be included in the map, then change one of the keys to be different:

const map = <int, String>{1: 'a', 2: 'b', 3: 'c', 4: 'd'};

If only one of the entries is needed, then remove the one that isn’t needed:

const map = <int, String>{1: 'a', 2: 'b', 4: 'd'};

Note that literal maps preserve the order of their entries, so the choice of which entry to remove might affect the order in which keys and values are returned by an iterator.

equal_keys_in_map

Two keys in a map literal shouldn’t be equal.

Description

The analyzer produces this diagnostic when a key in a non-constant map is the same as a previous key in the same map. If two keys are the same, then the second value overwrites the first value, which makes having both pairs pointless and likely signals a bug.

Example

The following code produces this diagnostic because the keys a and b have the same value:

const a = 1;
const b = 1;
var m = <int, String>{a: 'a', b: 'b'};

Common fixes

If both entries should be included in the map, then change one of the keys:

const a = 1;
const b = 2;
var m = <int, String>{a: 'a', b: 'b'};

If only one of the entries is needed, then remove the one that isn’t needed:

const a = 1;
var m = <int, String>{a: 'a'};

Note that literal maps preserve the order of their entries, so the choice of which entry to remove might affect the order in which the keys and values are returned by an iterator.

export_legacy_symbol

The symbol ‘{0}’ is defined in a legacy library, and can’t be re-exported from a library with null safety enabled.

Description

The analyzer produces this diagnostic when a library that was opted in to null safety exports another library, and the exported library is opted out of null safety.

Example

Given a library that is opted out of null safety:

// @dart = 2.8
String s;

The following code produces this diagnostic because it’s exporting symbols from an opted-out library:

export 'optedOut.dart';

class C {}

Common fixes

If you’re able to do so, migrate the exported library so that it doesn’t need to opt out:

String? s;

If you can’t migrate the library, then remove the export:

class C {}

If the exported library (the one that is opted out) itself exports an opted-in library, then it’s valid for your library to indirectly export the symbols from the opted-in library. You can do so by adding a hide combinator to the export directive in your library that hides all of the names declared in the opted-out library.

expression_in_map

Expressions can’t be used in a map literal.

Description

The analyzer produces this diagnostic when the analyzer finds an expression, rather than a map entry, in what appears to be a map literal.

Examples

The following code produces this diagnostic:

var map = <String, int>{'a': 0, 'b': 1, 'c'};

Common fixes

If the expression is intended to compute either a key or a value in an entry, fix the issue by replacing the expression with the key or the value. For example:

var map = <String, int>{'a': 0, 'b': 1, 'c': 2};

extends_non_class

Classes can only extend other classes.

Description

The analyzer produces this diagnostic when an extends clause contains a name that is declared to be something other than a class.

Examples

The following code produces this diagnostic because f is declared to be a function:

void f() {}

class C extends f {}

Common fixes

If you want the class to extend a class other than Object, then replace the name in the extends clause with the name of that class:

void f() {}

class C extends B {}

class B {}

If you want the class to extend Object, then remove the extends clause:

void f() {}

class C {}

extension_as_expression

Extension ‘{0}’ can’t be used as an expression.

Description

The analyzer produces this diagnostic when the name of an extension is used in an expression other than in an extension override or to qualify an access to a static member of the extension. Because classes define a type, the name of a class can be used to refer to the instance of Type representing the type of the class. Extensions, on the other hand, don’t define a type and can’t be used as a type literal.

Examples

The following code produces this diagnostic because E is an extension:

extension E on int {
  static String m() => '';
}

var x = E;

Common fixes

Replace the name of the extension with a name that can be referenced, such as a static member defined on the extension:

extension E on int {
  static String m() => '';
}

var x = E.m();

extension_conflicting_static_and_instance

Extension ‘{0}’ can’t define static member ‘{1}’ and an instance member with the same name.

Description

The analyzer produces this diagnostic when an extension declaration contains both an instance member and a static member that have the same name. The instance member and the static member can’t have the same name because it’s unclear which member is being referenced by an unqualified use of the name within the body of the extension.

Examples

The following code produces this diagnostic because the name a is being used for two different members:

extension E on Object {
  int get a => 0;
  static int a() => 0;
}

Common fixes

Rename or remove one of the members:

extension E on Object {
  int get a => 0;
  static int b() => 0;
}

extension_declares_abstract_member

Extensions can’t declare abstract members.

Description

The analyzer produces this diagnostic when an abstract declaration is declared in an extension. Extensions can declare only concrete members.

Examples

The following code produces this diagnostic because the method a doesn’t have a body:

extension E on String {
  int a();
}

Common fixes

Either provide an implementation for the member or remove it.

extension_declares_constructor

Extensions can’t declare constructors.

Description

The analyzer produces this diagnostic when a constructor declaration is found in an extension. It isn’t valid to define a constructor because extensions aren’t classes, and it isn’t possible to create an instance of an extension.

Examples

The following code produces this diagnostic because there is a constructor declaration in E:

extension E on String {
  E() : super();
}

Common fixes

Remove the constructor or replace it with a static method.

extension_declares_instance_field

Extensions can’t declare instance fields

Description

The analyzer produces this diagnostic when an instance field declaration is found in an extension. It isn’t valid to define an instance field because extensions can only add behavior, not state.

Examples

The following code produces this diagnostic because s is an instance field:

extension E on String {
  String s;
}

Common fixes

Remove the field, make it a static field, or convert it to be a getter, setter, or method.

extension_declares_member_of_object

Extensions can’t declare members with the same name as a member declared by ‘Object’.

Description

The analyzer produces this diagnostic when an extension declaration declares a member with the same name as a member declared in the class Object. Such a member can never be used because the member in Object is always found first.

Examples

The following code produces this diagnostic because toString is defined by Object:

extension E on String {
  String toString() => this;
}

Common fixes

Remove the member or rename it so that the name doesn’t conflict with the member in Object:

extension E on String {
  String displayString() => this;
}

extension_override_access_to_static_member

An extension override can’t be used to access a static member from an extension.

Description

The analyzer produces this diagnostic when an extension override is the receiver of the invocation of a static member. Similar to static members in classes, the static members of an extension should be accessed using the name of the extension, not an extension override.

Examples

The following code produces this diagnostic because m is static:

extension E on String {
  static void m() {}
}

void f() {
  E('').m();
}

Common fixes

Replace the extension override with the name of the extension:

extension E on String {
  static void m() {}
}

void f() {
  E.m();
}

extension_override_argument_not_assignable

The type of the argument to the extension override ‘{0}’ isn’t assignable to the extended type ‘{1}’.

Description

The analyzer produces this diagnostic when the argument to an extension override isn’t assignable to the type being extended by the extension.

Examples

The following code produces this diagnostic because 3 isn’t a String:

extension E on String {
  void method() {}
}

void f() {
  E(3).method();
}

Common fixes

If you’re using the correct extension, then update the argument to have the correct type:

extension E on String {
  void method() {}
}

void f() {
  E(3.toString()).method();
}

If there’s a different extension that’s valid for the type of the argument, then either replace the name of the extension or unwrap the argument so that the correct extension is found.

extension_override_without_access

An extension override can only be used to access instance members.

Description

The analyzer produces this diagnostic when an extension override is found that isn’t being used to access one of the members of the extension. The extension override syntax doesn’t have any runtime semantics; it only controls which member is selected at compile time.

Examples

The following code produces this diagnostic because E(i) isn’t an expression:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(E(i));
}

Common fixes

If you want to invoke one of the members of the extension, then add the invocation:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(E(i).a);
}

If you don’t want to invoke a member, then unwrap the argument:

extension E on int {
  int get a => 0;
}

void f(int i) {
  print(i);
}

extension_override_with_cascade

Extension overrides have no value so they can’t be used as the receiver of a cascade expression.

Description

The analyzer produces this diagnostic when an extension override is used as the receiver of a cascade expression. The value of a cascade expression e..m is the value of the receiver e, but extension overrides aren’t expressions and don’t have a value.

Examples

The following code produces this diagnostic because E(3) isn’t an expression:

extension E on int {
  void m() {}
}
f() {
  E(3)..m();
}

Common fixes

Use ‘.’ rather than ‘..’:

extension E on int {
  void m() {}
}
f() {
  E(3).m();
}

If there are multiple cascaded accesses, you’ll need to duplicate the extension override for each one.

extra_positional_arguments

Too many positional arguments: {0} expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has more positional arguments than the method or function allows.

Examples

The following code produces this diagnostic because f defines 2 parameters but is invoked with 3 arguments:

void f(int a, int b) {}
void g() {
  f(1, 2, 3);
}

Common fixes

Remove the arguments that don’t correspond to parameters:

void f(int a, int b) {}
void g() {
  f(1, 2);
}

extra_positional_arguments_could_be_named

Too many positional arguments: {0} expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has more positional arguments than the method or function allows, but the method or function defines named parameters.

Examples

The following code produces this diagnostic because f defines 2 positional parameters but has a named parameter that could be used for the third argument:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2, 3);
}

Common fixes

If some of the arguments should be values for named parameters, then add the names before the arguments:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2, c: 3);
}

Otherwise, remove the arguments that don’t correspond to positional parameters:

void f(int a, int b, {int c}) {}
void g() {
  f(1, 2);
}

field_initialized_in_initializer_and_declaration

Fields can’t be initialized in the constructor if they are final and were already initialized at their declaration.

Description

The analyzer produces this diagnostic when a final field is initialized in both the declaration of the field and in an initializer in a constructor. Final fields can only be assigned once, so it can’t be initialized in both places.

Example

The following code produces this diagnostic because f is :

class C {
  final int f = 0;
  C() : f = 1;
}

Common fixes

If the initialization doesn’t depend on any values passed to the constructor, and if all of the constructors need to initialize the field to the same value, then remove the initializer from the constructor:

class C {
  final int f = 0;
  C();
}

If the initialization depends on a value passed to the constructor, or if different constructors need to initialize the field differently, then remove the initializer in the field’s declaration:

class C {
  final int f;
  C() : f = 1;
}

field_initializer_not_assignable

The initializer type ‘{0}’ can’t be assigned to the field type ‘{1}’ in a const constructor.

The initializer type ‘{0}’ can’t be assigned to the field type ‘{1}’.

Description

The analyzer produces this diagnostic when the initializer list of a constructor initializes a field to a value that isn’t assignable to the field.

Example

The following code produces this diagnostic because 0 has the type int, and an int can’t be assigned to a field of type String:

class C {
  String s;

  C() : s = 0;
}

Common fixes

If the type of the field is correct, then change the value assigned to it so that the value has a valid type:

class C {
  String s;

  C() : s = '0';
}

If the type of the value is correct, then change the type of the field to allow the assignment:

class C {
  int s;

  C() : s = 0;
}

final_not_initialized

The final variable ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a final field or variable isn’t initialized.

Examples

The following code produces this diagnostic because x doesn’t have an initializer:

final x;

Common fixes

For variables and static fields, you can add an initializer:

final x = 0;

For instance fields, you can add an initializer as shown in the previous example, or you can initialize the field in every constructor. You can initialize the field by using a field formal parameter:

class C {
  final int x;
  C(this.x);
}

You can also initialize the field by using an initializer in the constructor:

class C {
  final int x;
  C(int y) : x = y * 2;
}

final_not_initialized_constructor

All final variables must be initialized, but ‘{0}’ and ‘{1}’ aren’t.

All final variables must be initialized, but ‘{0}’ isn’t.

All final variables must be initialized, but ‘{0}’, ‘{1}’, and {2} others aren’t.

Description

The analyzer produces this diagnostic when a class defines one or more final instance fields without initializers and has at least one constructor that doesn’t initialize those fields. All final instance fields must be initialized when the instance is created, either by the field’s initializer or by the constructor.

Examples

The following code produces this diagnostic:

class C {
  final String value;

  C();
}

Common fixes

If the value should be passed in to the constructor directly, then use a field formal parameter to initialize the field value:

class C {
  final String value;

  C(this.value);
}

If the value should be computed indirectly from a value provided by the caller, then add a parameter and include an initializer:

class C {
  final String value;

  C(Object o) : value = o.toString();
}

If the value of the field doesn’t depend on values that can be passed to the constructor, then add an initializer for the field as part of the field declaration:

class C {
  final String value = '';

  C();
}

If the value of the field doesn’t depend on values that can be passed to the constructor but different constructors need to initialize it to different values, then add an initializer for the field in the initializer list:

class C {
  final String value;

  C() : value = '';

  C.named() : value = 'c';
}

However, if the value is the same for all instances, then consider using a static field instead of an instance field:

class C {
  static const String value = '';

  C();
}

for_in_of_invalid_type

The type ‘{0}’ used in the ‘for’ loop must implement {1}.

Description

The analyzer produces this diagnostic when the expression following in in a for-in loop has a type that isn’t a subclass of Iterable.

Examples

The following code produces this diagnostic because m is a Map, and Map isn’t a subclass of Iterable:

void f(Map<String, String> m) {
  for (String s in m) {
    print(s);
  }
}

Common fixes

Replace the expression with one that produces an iterable value:

void f(Map<String, String> m) {
  for (String s in m.values) {
    print(s);
  }
}

getter_not_subtype_setter_types

The return type of getter ‘{0}’ is ‘{1}’ which isn’t a subtype of the type ‘{2}’ of its setter ‘{3}’.

Description

The analyzer produces this diagnostic when the return type of a getter isn’t a subtype of the type of the parameter of a setter with the same name.

The subtype relationship is a requirement whether the getter and setter are in the same class or whether one of them is in a superclass of the other.

Example

The following code produces this diagnostic because the return type of the getter x is num, the parameter type of the setter x is int, and num isn’t a subtype of int:

class C {
  num get x => 0;

  set x(int y) {}
}

Common fixes

If the type of the getter is correct, then change the type of the setter:

class C {
  num get x => 0;

  set x(num y) {}
}

If the type of the setter is correct, then change the type of the getter:

class C {
  int get x => 0;

  set x(int y) {}
}

illegal_async_return_type

Functions marked ‘async’ must have a return type assignable to ‘Future’.

Description

The analyzer produces this diagnostic when the body of a function has the async modifier even though the return type of the function isn’t assignable to Future.

Example

The following code produces this diagnostic because the body of the function f has the async modifier even though the return type isn’t assignable to Future:

int f() async {
  return 0;
}

Common fixes

If the function should be asynchronous, then change the return type to be assignable to Future:

Future<int> f() async {
  return 0;
}

If the function should be synchronous, then remove the async modifier:

int f() {
  return 0;
}

implements_non_class

Classes and mixins can only implement other classes and mixins.

Description

The analyzer produces this diagnostic when a name used in the implements clause of a class or mixin declaration is defined to be something other than a class or mixin.

Examples

The following code produces this diagnostic because x is a variable rather than a class or mixin:

var x;
class C implements x {}

Common fixes

If the name is the name of an existing class or mixin that’s already being imported, then add a prefix to the import so that the local definition of the name doesn’t shadow the imported name.

If the name is the name of an existing class or mixin that isn’t being imported, then add an import, with a prefix, for the library in which it’s declared.

Otherwise, either replace the name in the implements clause with the name of an existing class or mixin, or remove the name from the implements clause.

implements_repeated

‘{0}’ can only be implemented once.

Description

The analyzer produces this diagnostic when a single class is specified more than once in an implements clause.

Examples

The following code produces this diagnostic because A is in the list twice:

class A {}
class B implements A, A {}

Common fixes

Remove all except one occurrence of the class name:

class A {}
class B implements A {}

implicit_this_reference_in_initializer

The instance member ‘{0}’ can’t be accessed in an initializer.

Description

The analyzer produces this diagnostic when it finds a reference to an instance member in a constructor’s initializer list.

Examples

The following code produces this diagnostic because defaultX is an instance member:

class C {
  int x;

  C() : x = defaultX;

  int get defaultX => 0;
}

Common fixes

If the member can be made static, then do so:

class C {
  int x;

  C() : x = defaultX;

  static int get defaultX => 0;
}

If not, then replace the reference in the initializer with a different expression that doesn’t use an instance member:

class C {
  int x;

  C() : x = 0;

  int get defaultX => 0;
}

import_internal_library

The library ‘{0}’ is internal and can’t be imported.

Description

The analyzer produces this diagnostic when it finds an import whose dart: URI references an internal library.

Example

The following code produces this diagnostic because _interceptors is an internal library:

import 'dart:_interceptors';

Common fixes

Remove the import directive.

inconsistent_inheritance

Superinterfaces don’t have a valid override for ‘{0}’: {1}.

Description

The analyzer produces this diagnostic when a class inherits two or more conflicting signatures for a member and doesn’t provide an implementation that satisfies all the inherited signatures.

Example

The following code produces this diagnostic because C is inheriting the declaration of m from A, and that implementation isn’t consistent with the signature of m that’s inherited from B:

class A {
  void m({int a}) {}
}

class B {
  void m({int b}) {}
}

class C extends A implements B {
}

Common fixes

Add an implementation of the method that satisfies all the inherited signatures:

class A {
  void m({int a}) {}
}

class B {
  void m({int b}) {}
}

class C extends A implements B {
  void m({int a, int b}) {}
}

initializer_for_non_existent_field

‘{0}’ isn’t a field in the enclosing class.

Description

The analyzer produces this diagnostic when a constructor initializes a field that isn’t declared in the class containing the constructor. Constructors can’t initialize fields that aren’t declared and fields that are inherited from superclasses.

Examples

The following code produces this diagnostic because the initializer is initializing x, but x isn’t a field in the class:

class C {
  int y;

  C() : x = 0;
}

Common fixes

If a different field should be initialized, then change the name to the name of the field:

class C {
  int y;

  C() : y = 0;
}

If the field must be declared, then add a declaration:

class C {
  int x;
  int y;

  C() : x = 0;
}

initializing_formal_for_non_existent_field

‘{0}’ isn’t a field in the enclosing class.

Description

The analyzer produces this diagnostic when a field formal parameter is found in a constructor in a class that doesn’t declare the field being initialized. Constructors can’t initialize fields that aren’t declared and fields that are inherited from superclasses.

Examples

The following code produces this diagnostic because the field x isn’t defined:

class C {
  int y;

  C(this.x);
}

Common fixes

If the field name was wrong, then change it to the name of an existing field:

class C {
  int y;

  C(this.y);
}

If the field name is correct but hasn’t yet been defined, then declare the field:

class C {
  int x;
  int y;

  C(this.x);
}

If the parameter is needed but shouldn’t initialize a field, then convert it to a normal parameter and use it:

class C {
  int y;

  C(int x) : y = x * 2;
}

If the parameter isn’t needed, then remove it:

class C {
  int y;

  C();
}

instance_access_to_static_member

Static {1} ‘{0}’ can’t be accessed through an instance.

Description

The analyzer produces this diagnostic when an access operator is used to access a static member through an instance of the class.

Examples

The following code produces this diagnostic because zero is a static field, but it’s being accessed as if it were an instance field:

void f(C c) {
  c.zero;
}

class C {
  static int zero = 0;
}

Common fixes

Use the class to access the static member:

void f(C c) {
  C.zero;
}

class C {
  static int zero = 0;
}

instance_member_access_from_factory

Instance members can’t be accessed from a factory constructor.

Description

The analyzer produces this diagnostic when a factory constructor contains an unqualified reference to an instance member. In a generative constructor, the instance of the class is created and initialized before the body of the constructor is executed, so the instance can be bound to this and accessed just like it would be in an instance method. But, in a factory constructor, the instance isn’t created before executing the body, so this can’t be used to reference it.

Examples

The following code produces this diagnostic because x isn’t in scope in the factory constructor:

class C {
  int x;
  factory C() {
    return C._(x);
  }
  C._(this.x);
}

Common fixes

Rewrite the code so that it doesn’t reference the instance member:

class C {
  int x;
  factory C() {
    return C._(0);
  }
  C._(this.x);
}

instance_member_access_from_static

Instance members can’t be accessed from a static method.

Description

The analyzer produces this diagnostic when a static method contains an unqualified reference to an instance member.

Examples

The following code produces this diagnostic because the instance field x is being referenced in a static method:

class C {
  int x;

  static int m() {
    return x;
  }
}

Common fixes

If the method must reference the instance member, then it can’t be static, so remove the keyword:

class C {
  int x;

  int m() {
    return x;
  }
}

If the method can’t be made an instance method, then add a parameter so that an instance of the class can be passed in:

class C {
  int x;

  static int m(C c) {
    return c.x;
  }
}

instantiate_abstract_class

Abstract classes can’t be instantiated.

Description

The analyzer produces this diagnostic when it finds a constructor invocation and the constructor is declared in an abstract class. Even though you can’t create an instance of an abstract class, abstract classes can declare constructors that can be invoked by subclasses.

Examples

The following code produces this diagnostic because C is an abstract class:

abstract class C {}

var c = new C();

Common fixes

If there’s a concrete subclass of the abstract class that can be used, then create an instance of the concrete subclass.

invalid_annotation

Annotation must be either a const variable reference or const constructor invocation.

Getters can’t be used as annotations.

Description

The analyzer produces this diagnostic when an annotation is found that is using something that is neither a variable marked as const or the invocation of a const constructor.

Getters can’t be used as annotations.

Example

The following code produces this diagnostic because the variable v isn’t a const variable:

var v = 0;

@v
void f() {
}

The following code produces this diagnostic because f isn’t a variable:

@f
void f() {
}

The following code produces this diagnostic because f isn’t a constructor:

@f()
void f() {
}

The following code produces this diagnostic because g is a getter:

@g
int get g => 0;

Common fixes

If the annotation is referencing a variable that isn’t a const constructor, add the keyword const to the variable’s declaration:

const v = 0;

@v
void f() {
}

If the annotation isn’t referencing a variable, then remove it:

int v = 0;

void f() {
}

invalid_assignment

A value of type ‘{0}’ can’t be assigned to a variable of type ‘{1}’.

Description

The analyzer produces this diagnostic when the static type of an expression that is assigned to a variable isn’t assignable to the type of the variable.

Examples

The following code produces this diagnostic because the type of the initializer (int) isn’t assignable to the type of the variable (String):

int i = 0;
String s = i;

Common fixes

If the value being assigned is always assignable at runtime, even though the static types don’t reflect that, then add an explicit cast.

Otherwise, change the value being assigned so that it has the expected type. In the previous example, this might look like:

int i = 0;
String s = i.toString();

If you can’t change the value, then change the type of the variable to be compatible with the type of the value being assigned:

int i = 0;
int s = i;

invalid_extension_argument_count

Extension overrides must have exactly one argument: the value of ‘this’ in the extension method.

Description

The analyzer produces this diagnostic when an extension override doesn’t have exactly one argument. The argument is the expression used to compute the value of this within the extension method, so there must be one argument.

Examples

The following code produces this diagnostic because there are no arguments:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E().join('b');
}

And, the following code produces this diagnostic because there’s more than one argument:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E('a', 'b').join('c');
}

Common fixes

Provide one argument for the extension override:

extension E on String {
  String join(String other) => '$this $other';
}

void f() {
  E('a').join('b');
}

invalid_factory_name_not_a_class

The name of a factory constructor must be the same as the name of the immediately enclosing class.

Description

The analyzer produces this diagnostic when the name of a factory constructor isn’t the same as the name of the surrounding class.

Examples

The following code produces this diagnostic because the name of the factory constructor (A) isn’t the same as the surrounding class (C):

class A {}

class C {
  factory A() => throw 0;
}

Common fixes

If the factory returns an instance of the surrounding class, then rename the factory:

class A {}

class C {
  factory C() => throw 0;
}

If the factory returns an instance of a different class, then move the factory to that class:

class A {
  factory A() => throw 0;
}

class C {}

If the factory returns an instance of a different class, but you can’t modify that class or don’t want to move the factory, then convert it to be a static method:

class A {}

class C {
  static A a() => throw 0;
}

invalid_literal_annotation

Only const constructors can have the @literal annotation.

Description

The analyzer produces this diagnostic when the @literal annotation is applied to anything other than a const constructor.

Examples

The following code produces this diagnostic because the constructor isn’t a const constructor:

import 'package:meta/meta.dart';

class C {
  @literal
  C();
}

The following code produces this diagnostic because x isn’t a constructor:

import 'package:meta/meta.dart';

@literal
var x;

Common fixes

If the annotation is on a constructor and the constructor should always be invoked with const, when possible, then mark the constructor with the const keyword:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

If the constructor can’t be marked as const, then remove the annotation.

If the annotation is on anything other than a constructor, then remove the annotation:

var x;

invalid_null_aware_operator

The receiver can’t be null because of short-circuiting, so the null-aware operator ‘{0}’ can’t be used.

The receiver can’t be null, so the null-aware operator ‘{0}’ can’t be used.

Description

The analyzer produces this diagnostic when a null-aware operator (?., ?.., ?[, ?..[, or ...?) is used on a target that’s known to be non-nullable.

Example

The following code produces this diagnostic because s can’t be null:

int? getLength(String s) {
  return s?.length;
}

The following code produces this diagnostic because a can’t be null:

var a = [];
var b = [...?a];

The following code produces this diagnostic because s?.length can’t return null:

void f(String? s) {
  s?.length?.isEven;
}

The reason s?.length can’t return null is because the null-aware operator following s short-circuits the evaluation of both length and isEven if s is null. In other words, if s is null, then neither length nor isEven will be invoked, and if s is non-null, then length can’t return a null value. Either way, isEven can’t be invoked on a null value, so the null-aware operator is neither necessary nor allowed. See Understanding null safety for more details.

Common fixes

Replace the null-aware operator with a non-null-aware equivalent; for example, change ?. to .:

int getLength(String s) {
  return s.length;
}

(Note that the return type was also changed to be non-nullable, which might not be appropriate in some cases.)

invalid_override

‘{1}.{0}’ (‘{2}’) isn’t a valid override of ‘{3}.{0}’ (‘{4}’).

Description

The analyzer produces this diagnostic when a member of a class is found that overrides a member from a supertype and the override isn’t valid. An override is valid if all of these are true:

  • It allows all of the arguments allowed by the overridden member.
  • It doesn’t require any arguments that aren’t required by the overridden member.
  • The type of every parameter of the overridden member is assignable to the corresponding parameter of the override.
  • The return type of the override is assignable to the return type of the overridden member.

Examples

The following code produces this diagnostic because the type of the parameter s (String) isn’t assignable to the type of the parameter i (int):

class A {
  void m(int i) {}
}

class B extends A {
  void m(String s) {}
}

Common fixes

If the invalid method is intended to override the method from the superclass, then change it to conform:

class A {
  void m(int i) {}
}

class B extends A {
  void m(int i) {}
}

If it isn’t intended to override the method from the superclass, then rename it:

class A {
  void m(int i) {}
}

class B extends A {
  void m2(String s) {}
}

invalid_reference_to_this

Invalid reference to ‘this’ expression.

Description

The analyzer produces this diagnostic when this is used outside of an instance method or a generative constructor. The reserved word this is only defined in the context of an instance method or a generative constructor.

Examples

The following code produces this diagnostic because v is a top-level variable:

C f() => this;

class C {}

Common fixes

Use a variable of the appropriate type in place of this, declaring it if necessary:

C f(C c) => c;

class C {}

invalid_uri

Invalid URI syntax: ‘{0}’.

Description

The analyzer produces this diagnostic when a URI in a directive doesn’t conform to the syntax of a valid URI.

Examples

The following code produces this diagnostic because '#' isn’t a valid URI:

import '#';

Common fixes

Replace the invalid URI with a valid URI.

invalid_use_of_covariant_in_extension

Can’t have modifier ‘#lexeme’ in an extension.

Description

The analyzer produces this diagnostic when a member declared inside an extension uses the keyword covariant in the declaration of a parameter. Extensions aren’t classes and don’t have subclasses, so the keyword serves no purpose.

Examples

The following code produces this diagnostic because i is marked as being covariant:

extension E on String {
  void a(covariant int i) {}
}

Common fixes

Remove the ‘covariant’ keyword:

extension E on String {
  void a(int i) {}
}

invalid_use_of_null_value

An expression whose value is always ‘null’ can’t be dereferenced.

Description

The analyzer produces this diagnostic when an expression whose value will always be null is dererenced.

Example

The following code produces this diagnostic because x will always be null:

int f(Null x) {
  return x.length;
}

Common fixes

If the value is allowed to be something other than null, then change the type of the expression:

int f(String? x) {
  return x!.length;
}

invalid_visibility_annotation

The member ‘{0}’ is annotated with ‘{1}’, but this annotation is only meaningful on declarations of public members.

Description

The analyzer produces this diagnostic when either the @visibleForTemplate or @visibleForTesting annotation is applied to a non-public declaration.

Examples

The following code produces this diagnostic:

import 'package:meta/meta.dart';

@visibleForTesting
void _someFunction() {}

void f() => _someFunction();

Common fixes

If the declaration doesn’t need to be used by test code, then remove the annotation:

void _someFunction() {}

void f() => _someFunction();

If it does, then make it public:

import 'package:meta/meta.dart';

@visibleForTesting
void someFunction() {}

void f() => someFunction();

invocation_of_extension_without_call

The extension ‘{0}’ doesn’t define a ‘call’ method so the override can’t be used in an invocation.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a function but the extension doesn’t declare a call method.

Examples

The following code produces this diagnostic because the extension E doesn’t define a call method:

extension E on String {}

void f() {
  E('')();
}

Common fixes

If the extension is intended to define a call method, then declare it:

extension E on String {
  int call() => 0;
}

void f() {
  E('')();
}

If the extended type defines a call method, then remove the extension override.

If the call method isn’t defined, then rewrite the code so that it doesn’t invoke the call method.

invocation_of_non_function

‘{0}’ isn’t a function.

Description

The analyzer produces this diagnostic when it finds a function invocation, but the name of the function being invoked is defined to be something other than a function.

Examples

The following code produces this diagnostic because Binary is the name of a function type, not a function:

typedef Binary = int Function(int, int);

int f() {
  return Binary(1, 2);
}

Common fixes

Replace the name with the name of a function.

invocation_of_non_function_expression

The expression doesn’t evaluate to a function, so it can’t be invoked.

Description

The analyzer produces this diagnostic when a function invocation is found, but the name being referenced isn’t the name of a function, or when the expression computing the function doesn’t compute a function.

Examples

The following code produces this diagnostic because x isn’t a function:

int x = 0;

int f() => x;

var y = x();

The following code produces this diagnostic because f() doesn’t return a function:

int x = 0;

int f() => x;

var y = f()();

Common fixes

If you need to invoke a function, then replace the code before the argument list with the name of a function or with an expression that computes a function:

int x = 0;

int f() => x;

var y = f();

late_final_field_with_const_constructor

Can’t have a late final field in a class with a const constructor.

Description

The analyzer produces this diagnostic when a class that has at least one const constructor also has a field marked both late and final.

Example

The following code produces this diagnostic because the class A has a const constructor and the final field f is marked as late:

class A {
  late final int f;

  const A();
}

Common fixes

If the field doesn’t need to be marked late, then remove the late modifier from the field:

class A {
  final int f = 0;

  const A();
}

If the field must be marked late, then remove the const modifier from the constructors:

class A {
  late final int f;

  A();
}

late_final_local_already_assigned

The late final local variable is already assigned.

Description

The analyzer produces this diagnostic when the analyzer can prove that a local variable marked as both late and final was already assigned a value at the point where another assignment occurs.

Because final variables can only be assigned once, subsequent assignments are guaranteed to fail, so they’re flagged.

Example

The following code produces this diagnostic because the final variable v is assigned a value in two places:

int f() {
  late final int v;
  v = 0;
  v += 1;
  return v;
}

Common fixes

If you need to be able to reassign the variable, then remove the final keyword:

int f() {
  late int v;
  v = 0;
  v += 1;
  return v;
}

If you don’t need to reassign the variable, then remove all except the first of the assignments:

int f() {
  late final int v;
  v = 0;
  return v;
}

list_element_type_not_assignable

The element type ‘{0}’ can’t be assigned to the list type ‘{1}’.

Description

The analyzer produces this diagnostic when the type of an element in a list literal isn’t assignable to the element type of the list.

Examples

The following code produces this diagnostic because 2.5 is a double, and the list can hold only integers:

List<int> x = [1, 2.5, 3];

Common fixes

If you intended to add a different object to the list, then replace the element with an expression that computes the intended object:

List<int> x = [1, 2, 3];

If the object shouldn’t be in the list, then remove the element:

List<int> x = [1, 3];

If the object being computed is correct, then widen the element type of the list to allow all of the different types of objects it needs to contain:

List<num> x = [1, 2.5, 3];

map_entry_not_in_map

Map entries can only be used in a map literal.

Description

The analyzer produces this diagnostic when a map entry (a key/value pair) is found in a set literal.

Examples

The following code produces this diagnostic because the literal has a map entry even though it’s a set literal:

const collection = <String>{'a' : 'b'};

Common fixes

If you intended for the collection to be a map, then change the code so that it is a map. In the previous example, you could do this by adding another type argument:

const collection = <String, String>{'a' : 'b'};

In other cases, you might need to change the explicit type from Set to Map.

If you intended for the collection to be a set, then remove the map entry, possibly by replacing the colon with a comma if both values should be included in the set:

const collection = <String>{'a', 'b'};

map_key_type_not_assignable

The element type ‘{0}’ can’t be assigned to the map key type ‘{1}’.

Description

The analyzer produces this diagnostic when a key of a key-value pair in a map literal has a type that isn’t assignable to the key type of the map.

Examples

The following code produces this diagnostic because 2 is an int, but the keys of the map are required to be Strings:

var m = <String, String>{2 : 'a'};

Common fixes

If the type of the map is correct, then change the key to have the correct type:

var m = <String, String>{'2' : 'a'};

If the type of the key is correct, then change the key type of the map:

var m = <int, String>{2 : 'a'};

map_value_type_not_assignable

The element type ‘{0}’ can’t be assigned to the map value type ‘{1}’.

Description

The analyzer produces this diagnostic when a value of a key-value pair in a map literal has a type that isn’t assignable to the the value type of the map.

Examples

The following code produces this diagnostic because 2 is an int, but/ the values of the map are required to be Strings:

var m = <String, String>{'a' : 2};

Common fixes

If the type of the map is correct, then change the value to have the correct type:

var m = <String, String>{'a' : '2'};

If the type of the value is correct, then change the value type of the map:

var m = <String, int>{'a' : 2};

missing_default_value_for_parameter

The parameter ‘{0}’ can’t have a value of ‘null’ because of its type, and no non-null default value is provided.

Description

The analyzer produces this diagnostic when an optional parameter, whether positional or named, has a potentially non-nullable type and doesn’t specify a default value. Optional parameters that have no explicit default value have an implicit default value of null. If the type of the parameter doesn’t allow the parameter to have a value of null, then the implicit default value isn’t valid.

Example

The following code produces this diagnostic because x can’t be null, and no non-null default value is specified:

void f([int x]) {}

As does this:

void g({int x}) {}

Common fixes

If you want to use null to indicate that no value was provided, then you need to make the type nullable:

void f([int? x]) {}
void g({int? x}) {}

If the parameter can’t be null, then either provide a default value:

void f([int x = 1]) {}
void g({int x = 2}) {}

or make the parameter a required parameter:

void f(int x) {}
void g({required int x}) {}

missing_enum_constant_in_switch

Missing case clause for ‘{0}’.

Description

The analyzer produces this diagnostic when a switch statement for an enum doesn’t include an option for one of the values in the enumeration.

Note that null is always a possible value for an enum and therefore also must be handled.

Examples

The following code produces this diagnostic because the enum constant e2 isn’t handled:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
  }
}

Common fixes

If there’s special handling for the missing values, then add a case clause for each of the missing values:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
    case E.e2:
      break;
  }
}

If the missing values should be handled the same way, then add a default clause:

enum E { e1, e2 }

void f(E e) {
  switch (e) {
    case E.e1:
      break;
    default:
      break;
  }
}

missing_required_argument

The named parameter ‘{0}’ is required, but there’s no corresponding argument.

Description

The analyzer produces this diagnostic when an invocation of a function is missing a required named parameter.

Example

The following code produces this diagnostic because the invocation of f doesn’t include a value for the required named parameter end:

void f(int start, {required int end}) {}
void g() {
  f(3);
}

Common fixes

Add a named argument corresponding to the missing required parameter:

void f(int start, {required int end}) {}
void g() {
  f(3, end: 5);
}

missing_required_param

The parameter ‘{0}’ is required.

The parameter ‘{0}’ is required. {1}.

Description

The analyzer produces this diagnostic when a method or function with a named parameter that is annotated as being required is invoked without providing a value for the parameter.

Examples

The following code produces this diagnostic because the named parameter x is required:

import 'package:meta/meta.dart';

void f({@required int x}) {}

void g() {
  f();
}

Common fixes

Provide the required value:

import 'package:meta/meta.dart';

void f({@required int x}) {}

void g() {
  f(x: 2);
}

missing_return

This function has a return type of ‘{0}’, but doesn’t end with a return statement.

Description

Any function or method that doesn’t end with either an explicit return or a throw implicitly returns null. This is rarely the desired behavior. The analyzer produces this diagnostic when it finds an implicit return.

Examples

The following code produces this diagnostic because f doesn’t end with a return:

int f(int x) {
  if (x < 0) {
    return 0;
  }
}

Common fixes

Add a return statement that makes the return value explicit, even if null is the appropriate value.

mixin_of_non_class

Classes can only mix in mixins and classes.

Description

The analyzer produces this diagnostic when a name in a with clause is defined to be something other than a mixin or a class.

Examples

The following code produces this diagnostic because F is defined to be a function type:

typedef F = int Function(String);

class C with F {}

Common fixes

Remove the invalid name from the list, possibly replacing it with the name of the intended mixin or class:

typedef F = int Function(String);

class C {}

mixin_on_sealed_class

The class ‘{0}’ shouldn’t be used as a mixin constraint because it is sealed, and any class mixing in this mixin must have ‘{0}’ as a superclass.

Description

The analyzer produces this diagnostic when the superclass constraint of a mixin is a class from a different package that was marked as @sealed. Classes that are sealed can’t be extended, implemented, mixed in, or used as a superclass constraint.

Examples

If the package ‘p’ defines a sealed class:

import 'package:meta/meta.dart';

@sealed
class C {}

Then, the following code, when in a package other than ‘p’, produces this diagnostic:

import 'package:p/p.dart';

mixin M on C {}

Common fixes

If the classes that use the mixin don’t need to be subclasses of the sealed class, then consider adding a field and delegating to the wrapped instance of the sealed class.

mixin_super_class_constraint_non_interface

Only classes and mixins can be used as superclass constraints.

Description

The analyzer produces this diagnostic when a type following the on keyword in a mixin declaration is neither a class nor a mixin.

Examples

The following code produces this diagnostic because F is neither a class nor a mixin:

typedef F = void Function();

mixin M on F {}

Common fixes

If the type was intended to be a class but was mistyped, then replace the name.

Otherwise, remove the type from the on clause.

must_be_immutable

This class (or a class that this class inherits from) is marked as ‘@immutable’, but one or more of its instance fields aren’t final: {0}

Description

The analyzer produces this diagnostic when an immutable class defines one or more instance fields that aren’t final. A class is immutable if it’s marked as being immutable using the annotation @immutable or if it’s a subclass of an immutable class.

Examples

The following code produces this diagnostic because the field x isn’t final:

import 'package:meta/meta.dart';

@immutable
class C {
  int x;

  C(this.x);
}

Common fixes

If instances of the class should be immutable, then add the keyword final to all non-final field declarations:

import 'package:meta/meta.dart';

@immutable
class C {
  final int x;

  C(this.x);
}

If the instances of the class should be mutable, then remove the

class C {
  int x;

  C(this.x);
}

must_call_super

This method overrides a method annotated as ‘@mustCallSuper’ in ‘{0}’, but doesn’t invoke the overridden method.

Description

The analyzer produces this diagnostic when a method that overrides a method that is annotated as @mustCallSuper doesn’t invoke the overridden method as required.

Examples

The following code produces this diagnostic because the method m in B doesn’t invoke the overridden method m in A:

import 'package:meta/meta.dart';

class A {
  @mustCallSuper
  m() {}
}

class B extends A {
  @override
  m() {}
}

Common fixes

Add an invocation of the overridden method in the overriding method:

import 'package:meta/meta.dart';

class A {
  @mustCallSuper
  m() {}
}

class B extends A {
  @override
  m() {
    super.m();
  }
}

new_with_undefined_constructor_default

The class ‘{0}’ doesn’t have a default constructor.

Description

The analyzer produces this diagnostic when an unnamed constructor is invoked on a class that defines named constructors but the class doesn’t have an unnamed constructor.

Examples

The following code produces this diagnostic because A doesn’t define an unnamed constructor:

class A {
  A.a();
}

A f() => A();

Common fixes

If one of the named constructors does what you need, then use it:

class A {
  A.a();
}

A f() => A.a();

If none of the named constructors does what you need, and you’re able to add an unnamed constructor, then add the constructor:

class A {
  A();
  A.a();
}

A f() => A();

non_abstract_class_inherits_abstract_member

Missing concrete implementation of ‘{0}’.

Missing concrete implementations of ‘{0}’ and ‘{1}’.

Missing concrete implementations of ‘{0}’, ‘{1}’, ‘{2}’, ‘{3}’, and {4} more.

Missing concrete implementations of ‘{0}’, ‘{1}’, ‘{2}’, and ‘{3}’.

Missing concrete implementations of ‘{0}’, ‘{1}’, and ‘{2}’.

Description

The analyzer produces this diagnostic when a concrete class inherits one or more abstract members, and doesn’t provide or inherit an implementation for at least one of those abstract members.

Examples

The following code produces this diagnostic because the class B doesn’t have a concrete implementation of m:

abstract class A {
  void m();
}

class B extends A {}

Common fixes

If the subclass can provide a concrete implementation for some or all of the abstract inherited members, then add the concrete implementations:

abstract class A {
  void m();
}

class B extends A {
  void m() {}
}

If there is a mixin that provides an implementation of the inherited methods, then apply the mixin to the subclass:

abstract class A {
  void m();
}

class B extends A with M {}

mixin M {
  void m() {}
}

If the subclass can’t provide a concrete implementation for all of the abstract inherited members, then mark the subclass as being abstract:

abstract class A {
  void m();
}

abstract class B extends A {}

non_bool_condition

Conditions must have a static type of ‘bool’.

Description

The analyzer produces this diagnostic when a condition, such as an if or while loop, doesn’t have the static type bool.

Examples

The following code produces this diagnostic because x has the static type int:

void f(int x) {
  if (x) {
    // ...
  }
}

Common fixes

Change the condition so that it produces a Boolean value:

void f(int x) {
  if (x == 0) {
    // ...
  }
}

non_bool_expression

The expression in an assert must be of type ‘bool’.

Description

The analyzer produces this diagnostic when the first expression in an assert has a type other than bool.

Examples

The following code produces this diagnostic because the type of p is int, but a bool is required:

void f(int p) {
  assert(p);
}

Common fixes

Change the expression so that it has the type bool:

void f(int p) {
  assert(p > 0);
}

non_bool_negation_expression

A negation operand must have a static type of ‘bool’.

Description

The analyzer produces this diagnostic when the operand of the unary negation operator (!) doesn’t have the type bool.

Examples

The following code produces this diagnostic because x is an int when it must be a bool:

int x = 0;
bool y = !x;

Common fixes

Replace the operand with an expression that has the type bool:

int x = 0;
bool y = !(x > 0);

non_bool_operand

The operands of the operator ‘{0}’ must be assignable to ‘bool’.

Description

The analyzer produces this diagnostic when one of the operands of either the && or || operator doesn’t have the type bool.

Examples

The following code produces this diagnostic because a isn’t a Boolean value:

int a = 3;
bool b = a || a > 1;

Common fixes

Change the operand to a Boolean value:

int a = 3;
bool b = a == 0 || a > 1;

non_constant_annotation_constructor

Annotation creation can only call a const constructor.

Description

The analyzer produces this diagnostic when an annotation is the invocation of an existing constructor even though the invoked constructor isn’t a const constructor.

Example

The following code produces this diagnostic because the constructor for C isn’t a const constructor:

@C()
void f() {
}

class C {
  C();
}

Common fixes

If it’s valid for the class to have a const constructor, then create a const constructor that can be used for the annotation:

@C()
void f() {
}

class C {
  const C();
}

If it isn’t valid for the class to have a const constructor, then either remove the annotation or use a different class for the annotation.

non_constant_case_expression

Case expressions must be constant.

Description

The analyzer produces this diagnostic when the expression in a case clause isn’t a constant expression.

Examples

The following code produces this diagnostic because j isn’t a constant:

void f(int i, int j) {
  switch (i) {
    case j:
      // ...
      break;
  }
}

Common fixes

Either make the expression a constant expression, or rewrite the switch statement as a sequence of if statements:

void f(int i, int j) {
  if (i == j) {
    // ...
  }
}

non_constant_default_value

The default value of an optional parameter must be constant.

Description

The analyzer produces this diagnostic when an optional parameter, either named or positional, has a default value that isn’t a compile-time constant.

Examples

The following code produces this diagnostic:

var defaultValue = 3;

void f([int value = defaultValue]) {}

Common fixes

If the default value can be converted to be a constant, then convert it:

const defaultValue = 3;

void f([int value = defaultValue]) {}

If the default value needs to change over time, then apply the default value inside the function:

var defaultValue = 3;

void f([int value]) {
  value ??= defaultValue;
}

non_constant_list_element

The values in a const list literal must be constants.

Description

The analyzer produces this diagnostic when an element in a constant list literal isn’t a constant value. The list literal can be constant either explicitly (because it’s prefixed by the const keyword) or implicitly (because it appears in a constant context).

Examples

The following code produces this diagnostic because x isn’t a constant, even though it appears in an implicitly constant list literal:

var x = 2;
var y = const <int>[0, 1, x];

Common fixes

If the list needs to be a constant list, then convert the element to be a constant. In the example above, you might add the const keyword to the declaration of x:

const x = 2;
var y = const <int>[0, 1, x];

If the expression can’t be made a constant, then the list can’t be a constant either, so you must change the code so that the list isn’t a constant. In the example above this means removing the const keyword before the list literal:

var x = 2;
var y = <int>[0, 1, x];

non_constant_map_element

The elements in a const map literal must be constant.

Description

The analyzer produces this diagnostic when an if element or a spread element in a constant map isn’t a constant element.

Examples

The following code produces this diagnostic because it’s attempting to spread a non-constant map:

var notConst = <int, int>{};
var map = const <int, int>{...notConst};

Similarly, the following code produces this diagnostic because the condition in the if element isn’t a constant expression:

bool notConst = true;
var map = const <int, int>{if (notConst) 1 : 2};

Common fixes

If the map needs to be a constant map, then make the elements constants. In the spread example, you might do that by making the collection being spread a constant:

const notConst = <int, int>{};
var map = const <int, int>{...notConst};

If the map doesn’t need to be a constant map, then remove the const keyword:

bool notConst = true;
var map = <int, int>{if (notConst) 1 : 2};

non_constant_map_key

The keys in a const map literal must be constant.

Description

The analyzer produces this diagnostic when a key in a constant map literal isn’t a constant value.

Examples

The following code produces this diagnostic beause a isn’t a constant:

var a = 'a';
var m = const {a: 0};

Common fixes

If the map needs to be a constant map, then make the key a constant:

const a = 'a';
var m = const {a: 0};

If the map doesn’t need to be a constant map, then remove the const keyword:

var a = 'a';
var m = {a: 0};

non_constant_map_value

The values in a const map literal must be constant.

Description

The analyzer produces this diagnostic when a value in a constant map literal isn’t a constant value.

Examples

The following code produces this diagnostic because a isn’t a constant:

var a = 'a';
var m = const {0: a};

Common fixes

If the map needs to be a constant map, then make the key a constant:

const a = 'a';
var m = const {0: a};

If the map doesn’t need to be a constant map, then remove the const keyword:

var a = 'a';
var m = {0: a};

non_constant_set_element

The values in a const set literal must be constants.

Description

The analyzer produces this diagnostic when a constant set literal contains an element that isn’t a compile-time constant.

Examples

The following code produces this diagnostic because i isn’t a constant:

var i = 0;

var s = const {i};

Common fixes

If the element can be changed to be a constant, then change it:

const i = 0;

var s = const {i};

If the element can’t be a constant, then remove the keyword const:

var i = 0;

var s = {i};

non_const_call_to_literal_constructor

This instance creation must be ‘const’, because the {0} constructor is marked as ‘@literal’.

This instance creation must be ‘const’, because the {0} constructor is marked as ‘@literal’.

Description

The analyzer produces this diagnostic when a constructor that has the @literal annotation is invoked without using the const keyword, but all of the arguments to the constructor are constants. The annotation indicates that the constructor should be used to create a constant value whenever possible.

Examples

The following code produces this diagnostic:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

C f() => C();

Common fixes

Add the keyword const before the constructor invocation:

import 'package:meta/meta.dart';

class C {
  @literal
  const C();
}

void f() => const C();

non_type_as_type_argument

The name ‘{0}’ isn’t a type so it can’t be used as a type argument.

Description

The analyzer produces this diagnostic when an identifier that isn’t a type is used as a type argument.

Examples

The following code produces this diagnostic because x is a variable, not a type:

var x = 0;
List<x> xList = [];

Common fixes

Change the type argument to be a type:

var x = 0;
List<int> xList = [];

non_type_in_catch_clause

The name ‘{0}’ isn’t a type and can’t be used in an on-catch clause.

Description

The analyzer produces this diagnostic when the identifier following the on in a catch clause is defined to be something other than a type.

Examples

The following code produces this diagnostic because f is a function, not a type:

void f() {
  try {
    // ...
  } on f {
    // ...
  }
}

Common fixes

Change the name to the type of object that should be caught:

void f() {
  try {
    // ...
  } on FormatException {
    // ...
  }
}

not_assigned_potentially_non_nullable_local_variable

The non-nullable local variable ‘{0}’ must be assigned before it can be used.

Description

The analyzer produces this diagnostic when a local variable is referenced and has all these characteristics:

  • Has a type that’s potentially non-nullable.
  • Doesn’t have an initializer.
  • Isn’t marked as late.
  • The analyzer can’t prove that the local variable will be assigned before the reference based on the specification of [definite assignment.][]

Example

The following code produces this diagnostic because x can’t have a value of null, but is referenced before a value was assigned to it:

String f() {
  int x;
  return x.toString();
}

The following code produces this diagnostic because the assignment to x might not be executed, so it might have a value of null:

int g(bool b) {
  int x;
  if (b) {
    x = 1;
  }
  return x * 2;
}

The following code produces this diagnostic because the analyzer can’t prove, based on definite assignment analysis, that x won’t be referenced without having a value assigned to it:

int h(bool b) {
  int x;
  if (b) {
    x = 1;
  }
  if (b) {
    return x * 2;
  }
  return 0;
}

Common fixes

If null is a valid value, then make the variable nullable:

String f() {
  int? x;
  return x!.toString();
}

If null isn’t a valid value, and there’s a reasonable default value, then add an initializer:

int g(bool b) {
  int x = 2;
  if (b) {
    x = 1;
  }
  return x * 2;
}

Otherwise, ensure that a value was assigned on every possible code path before the value is accessed:

int g(bool b) {
  int x;
  if (b) {
    x = 1;
  } else {
    x = 2;
  }
  return x * 2;
}

You can also mark the variable as late, which removes the diagnostic, but if the variable isn’t assigned a value before it’s accessed, then it results in an exception being thrown at runtime. This approach should only be used if you’re sure that the variable will always be assigned, even though the analyzer can’t prove it based on definite assignment analysis.

int h(bool b) {
  late int x;
  if (b) {
    x = 1;
  }
  if (b) {
    return x * 2;
  }
  return 0;
}

not_a_type

{0} isn’t a type.

Description

The analyzer produces this diagnostic when a name is used as a type but declared to be something other than a type.

Examples

The following code produces this diagnostic because f is a function:

f() {}
g(f v) {}

Common fixes

Replace the name with the name of a type.

not_enough_positional_arguments

{0} positional argument(s) expected, but {1} found.

Description

The analyzer produces this diagnostic when a method or function invocation has fewer positional arguments than the number of required positional parameters.

Examples

The following code produces this diagnostic because f declares two required parameters, but only one argument is provided:

void f(int a, int b) {}
void g() {
  f(0);
}

Common fixes

Add arguments corresponding to the remaining parameters:

void f(int a, int b) {}
void g() {
  f(0, 1);
}

not_initialized_non_nullable_instance_field

Non-nullable instance field ‘{0}’ must be initialized.

Non-nullable instance field ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a field is declared and has all these characteristics:

Example

The following code produces this diagnostic because x is implicitly initialized to null when it isn’t allowed to be null:

class C {
  int x;
}

Similarly, the following code produces this diagnostic because x is implicitly initialized to null, when it isn’t allowed to be null, by one of the constructors, even though it’s initialized by other constructors:

class C {
  int x;

  C(this.x);

  C.n();
}

Common fixes

If there’s a reasonable default value for the field that’s the same for all instances, then add an initializer expression:

class C {
  int x = 0;
}

If the value of the field should be provided when an instance is created, then add a constructor that sets the value of the field or update an existing constructor:

class C {
  int x;

  C(this.x);
}

You can also mark the field as late, which removes the diagnostic, but if the field isn’t assigned a value before it’s accessed, then it results in an exception being thrown at runtime. This approach should only be used if you’re sure that the field will always be assigned before it’s referenced.

class C {
  late int x;
}

not_initialized_non_nullable_variable

The non-nullable variable ‘{0}’ must be initialized.

Description

The analyzer produces this diagnostic when a static field or top-level variable has a type that’s non-nullable and doesn’t have an initializer. Fields and variables that don’t have an initializer are normally initialized to null, but the type of the field or variable doesn’t allow it to be set to null, so an explicit initializer must be provided.

Example

The following code produces this diagnostic because the field f can’t be initialized to null:

class C {
  static int f;
}

Similarly, the following code produces this diagnostic because the top-level variable v can’t be initialized to null:

int v;

Common fixes

If the field or variable can’t be initialized to null, then add an initializer that sets it to a non-null value:

class C {
  static int f = 0;
}

If the field or variable should be initialized to null, then change the type to be nullable:

int? v;

If the field or variable can’t be initialized in the declaration but will always be initialized before it’s referenced, then mark it as being late:

class C {
  static late int f;
}

not_iterable_spread

Spread elements in list or set literals must implement ‘Iterable’.

Description

The analyzer produces this diagnostic when the static type of the expression of a spread element that appears in either a list literal or a set literal doesn’t implement the type Iterable.

Examples

The following code produces this diagnostic:

var m = <String, int>{'a': 0, 'b': 1};
var s = <String>{...m};

Common fixes

The most common fix is to replace the expression with one that produces an iterable object:

var m = <String, int>{'a': 0, 'b': 1};
var s = <String>{...m.keys};

not_map_spread

Spread elements in map literals must implement ‘Map’.

Description

The analyzer produces this diagnostic when the static type of the expression of a spread element that appears in a map literal doesn’t implement the type Map.

Examples

The following code produces this diagnostic because l isn’t a Map:

var l =  <String>['a', 'b'];
var m = <int, String>{...l};

Common fixes

The most common fix is to replace the expression with one that produces a map:

var l =  <String>['a', 'b'];
var m = <int, String>{...l.asMap()};

no_annotation_constructor_arguments

Annotation creation must have arguments.

Description

The analyzer produces this diagnostic when an annotation consists of a single identifier, but that identifier is the name of a class rather than a variable. To create an instance of the class, the identifier must be followed by an argument list.

Examples

The following code produces this diagnostic because C is a class, and a class can’t be used as an annotation without invoking a const constructor from the class:

class C {
  const C();
}

@C
var x;

Common fixes

Add the missing argument list:

class C {
  const C();
}

@C()
var x;

no_combined_super_signature

Can’t infer missing types in ‘{0}’ from overridden methods: {1}.

Description

The analyzer produces this diagnostic when there is a method declaration for which one or more types needs to be inferred, and those types can’t be inferred because none of the overridden methods has a function type that is a supertype of all the other overridden methods, as specified by override inference.

Example

The following code produces this diagnostic because the method m declared in the class C is missing both the return type and the type of the parameter a, and neither of the missing types can be inferred for it:

abstract class A {
  A m(String a);
}

abstract class B {
  B m(int a);
}

abstract class C implements A, B {
  m(a);
}

In this example, override inference can’t be performed because the overridden methods are incompatible in these ways:

  • Neither parameter type (String and int) is a supertype of the other.
  • Neither return type is a subtype of the other.

Common fixes

If possible, add types to the method in the subclass that are consistent with the types from all the overridden methods:

abstract class A {
  A m(String a);
}

abstract class B {
  B m(int a);
}

abstract class C implements A, B {
  C m(Object a);
}

nullable_type_in_catch_clause

A potentially nullable type can’t be used in an ‘on’ clause because it isn’t valid to throw a nullable expression.

Description

The analyzer produces this diagnostic when the type following on in a catch clause is a nullable type. It isn’t valid to specify a nullable type because it isn’t possible to catch null (because it’s a runtime error to throw null).

Example

The following code produces this diagnostic because the exception type is specified to allow null when null can’t be thrown:

void f() {
  try {
    // ...
  } on FormatException? {
  }
}

Common fixes

Remove the question mark from the type:

void f() {
  try {
    // ...
  } on FormatException {
  }
}

nullable_type_in_extends_clause

A class can’t extend a nullable type.

Description

The analyzer produces this diagnostic when a class declaration uses an extends clause to specify a superclass, and the superclass is followed by a ?.

It isn’t valid to specify a nullable superclass because doing so would have no meaning; it wouldn’t change either the interface or implementation being inherited by the class containing the extends clause.

Note, however, that it is valid to use a nullable type as a type argument to the superclass, such as class A extends B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in an extends clause:

class A {}
class B extends A? {}

Common fixes

Remove the question mark from the type:

class A {}
class B extends A {}

nullable_type_in_implements_clause

A class or mixin can’t implement a nullable type.

Description

The analyzer produces this diagnostic when a class or mixin declaration has an implements clause, and an interface is followed by a ?.

It isn’t valid to specify a nullable interface because doing so would have no meaning; it wouldn’t change the interface being inherited by the class containing the implements clause.

Note, however, that it is valid to use a nullable type as a type argument to the interface, such as class A implements B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in an implements clause:

class A {}
class B implements A? {}

Common fixes

Remove the question mark from the type:

class A {}
class B implements A {}

nullable_type_in_on_clause

A mixin can’t have a nullable type as a superclass constraint.

Description

The analyzer produces this diagnostic when a mixin declaration uses an on clause to specify a superclass constraint, and the class that’s specified is followed by a ?.

It isn’t valid to specify a nullable superclass constraint because doing so would have no meaning; it wouldn’t change the interface being depended on by the mixin containing the on clause.

Note, however, that it is valid to use a nullable type as a type argument to the superclass constraint, such as mixin A on B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type and nullable types can’t be used in an on clause:

class C {}
mixin M on C? {}

Common fixes

Remove the question mark from the type:

class C {}
mixin M on C {}

nullable_type_in_with_clause

A class or mixin can’t mix in a nullable type.

Description

The analyzer produces this diagnostic when a class or mixin declaration has a with clause, and a mixin is followed by a ?.

It isn’t valid to specify a nullable mixin because doing so would have no meaning; it wouldn’t change either the interface or implementation being inherited by the class containing the with clause.

Note, however, that it is valid to use a nullable type as a type argument to the mixin, such as class A with B<C?> {}.

Example

The following code produces this diagnostic because A? is a nullable type, and nullable types can’t be used in a with clause:

mixin M {}
class C with M? {}

Common fixes

Remove the question mark from the type:

mixin M {}
class C with M {}

override_on_non_overriding_member

The field doesn’t override an inherited getter or setter.

The getter doesn’t override an inherited getter.

The method doesn’t override an inherited method.

The setter doesn’t override an inherited setter.

Description

The analyzer produces this diagnostic when a class member is annotated with the @override annotation, but the member isn’t declared in any of the supertypes of the class.

Examples

The following code produces this diagnostic because m isn’t declared in any of the supertypes of C:

class C {
  @override
  String m() => '';
}

Common fixes

If the member is intended to override a member with a different name, then update the member to have the same name:

class C {
  @override
  String toString() => '';
}

If the member is intended to override a member that was removed from the superclass, then consider removing the member from the subclass.

If the member can’t be removed, then remove the annotation.

part_of_different_library

Expected this library to be part of ‘{0}’, not ‘{1}’.

Description

The analyzer produces this diagnostic when a library attempts to include a file as a part of itself when the other file is a part of a different library.

Example

Given a file named part.dart containing

part of 'library.dart';

The following code, in any file other than library.dart, produces this diagnostic because it attempts to include part.dart as a part of itself when part.dart is a part of a different library:

part 'package:a/part.dart';

Common fixes

If the library should be using a different file as a part, then change the URI in the part directive to be the URI of the other file.

If the part file should be a part of this library, then update the URI (or library name) in the part-of directive to be the URI (or name) of the correct library.

part_of_non_part

The included part ‘{0}’ must have a part-of directive.

Description

The analyzer produces this diagnostic when a part directive is found and the referenced file doesn’t have a part-of directive.

Examples

Given a file (a.dart) containing:

class A {}

The following code produces this diagnostic because a.dart doesn’t contain a part-of directive:

part 'a.dart';

Common fixes

If the referenced file is intended to be a part of another library, then add a part-of directive to the file:

part of 'test.dart';

class A {}

If the referenced file is intended to be a library, then replace the part directive with an import directive:

import 'a.dart';

prefix_collides_with_top_level_member

The name ‘{0}’ is already used as an import prefix and can’t be used to name a top-level element.

Description

The analyzer produces this diagnostic when a name is used as both an import prefix and the name of a top-level declaration in the same library.

Example

The following code produces this diagnostic because f is used as both an import prefix and the name of a function:

import 'dart:math' as f;

int f() => f.min(0, 1);

Common fixes

If you want to use the name for the import prefix, then rename the top-level declaration:

import 'dart:math' as f;

int g() => f.min(0, 1);

If you want to use the name for the top-level declaration, then rename the import prefix:

import 'dart:math' as math;

int f() => math.min(0, 1);

prefix_identifier_not_followed_by_dot

The name ‘{0}’ refers to an import prefix, so it must be followed by ‘.’.

Description

The analyzer produces this diagnostic when an import prefix is used by itself, without accessing any of the names declared in the libraries associated with the prefix. Prefixes aren’t variables, and therefore can’t be used as a value.

Example

The following code produces this diagnostic because the prefix math is being used as if it were a variable:

import 'dart:math' as math;

void f() {
  print(math);
}

Common fixes

If the code is incomplete, then reference something in one of the libraries associated with the prefix:

import 'dart:math' as math;

void f() {
  print(math.pi);
}

If the name is wrong, then correct the name.

redirect_to_invalid_function_type

The redirected constructor ‘{0}’ has incompatible parameters with ‘{1}’.

Description

The analyzer produces this diagnostic when a factory constructor attempts to redirect to another constructor, but the two have incompatible parameters. The parameters are compatible if all of the parameters of the redirecting constructor can be passed to the other constructor and if the other constructor doesn’t require any parameters that aren’t declared by the redirecting constructor.

Examples

The following code produces this diagnostic because the constructor for A doesn’t declare a parameter that the constructor for B requires:

abstract class A {
  factory A() = B;
}

class B implements A {
  B(int x);
  B.zero();
}

The following code produces this diagnostic because the constructor for A declares a named parameter (y) that the constructor for B doesn’t allow:

abstract class A {
  factory A(int x, {int y}) = B;
}

class B implements A {
  B(int x);
}

Common fixes

If there’s a different constructor that is compatible with the redirecting constructor, then redirect to that constructor:

abstract class A {
  factory A() = B.zero;
}

class B implements A {
  B(int x);
  B.zero();
}

Otherwise, update the redirecting constructor to be compatible:

abstract class A {
  factory A(int x) = B;
}

class B implements A {
  B(int x);
}

redirect_to_invalid_return_type

The return type ‘{0}’ of the redirected constructor isn’t a subtype of ‘{1}’.

Description

The analyzer produces this diagnostic when a factory constructor redirects to a constructor whose return type isn’t a subtype of the type that the factory constructor is declared to produce.

Examples

The following code produces this diagnostic because A isn’t a subclass of C, which means that the value returned by the constructor A() couldn’t be returned from the constructor C():

class A {}

class B implements C {}

class C {
  factory C() = A;
}

Common fixes

If the factory constructor is redirecting to a constructor in the wrong class, then update the factory constructor to redirect to the correct constructor:

class A {}

class B implements C {}

class C {
  factory C() = B;
}

If the class defining the constructor being redirected to is the class that should be returned, then make it a subtype of the factory’s return type:

class A implements C {}

class B implements C {}

class C {
  factory C() = A;
}

redirect_to_non_class

The name ‘{0}’ isn’t a type and can’t be used in a redirected constructor.

Description

One way to implement a factory constructor is to redirect to another constructor by referencing the name of the constructor. The analyzer produces this diagnostic when the redirect is to something other than a constructor.

Examples

The following code produces this diagnostic because f is a function:

C f() => throw 0;

class C {
  factory C() = f;
}

Common fixes

If the constructor isn’t defined, then either define it or replace it with a constructor that is defined.

If the constructor is defined but the class that defines it isn’t visible, then you probably need to add an import.

If you’re trying to return the value returned by a function, then rewrite the constructor to return the value from the constructor’s body:

C f() => throw 0;

class C {
  factory C() => f();
}

referenced_before_declaration

Local variable ‘{0}’ can’t be referenced before it is declared.

Description

The analyzer produces this diagnostic when a variable is referenced before it’s declared. In Dart, variables are visible everywhere in the block in which they are declared, but can only be referenced after they are declared.

The analyzer also produces a context message that indicates where the declaration is located.

Examples

The following code produces this diagnostic because i is used before it is declared:

void f() {
  print(i);
  int i = 5;
}

Common fixes

If you intended to reference the local variable, move the declaration before the first reference:

void f() {
  int i = 5;
  print(i);
}

If you intended to reference a name from an outer scope, such as a parameter, instance field or top-level variable, then rename the local declaration so that it doesn’t hide the outer variable.

void f(int i) {
  print(i);
  int x = 5;
  print(x);
}

return_in_generative_constructor

Constructors can’t return values.

Description

The analyzer produces this diagnostic when a generative constructor contains a return statement that specifies a value to be returned. Generative constructors always return the object that was created, and therefore can’t return a different object.

Example

The following code produces this diagnostic because the return statement has an expression:

class C {
  C() {
    return this;
  }
}

Common fixes

If the constructor should create a new instance, then remove either the return statement or the expression:

class C {
  C();
}

If the constructor shouldn’t create a new instance, then convert it to be a factory constructor:

class C {
  factory C() {
    return _instance;
  }

  static C _instance = C._();

  C._();
}

return_of_invalid_type

A value of type ‘{0}’ can’t be returned from constructor ‘{2}’ because it has a return type of ‘{1}’.

A value of type ‘{0}’ can’t be returned from function ‘{2}’ because it has a return type of ‘{1}’.

A value of type ‘{0}’ can’t be returned from method ‘{2}’ because it has a return type of ‘{1}’.

Description

The analyzer produces this diagnostic when a method or function returns a value whose type isn’t assignable to the declared return type.

Examples

The following code produces this diagnostic because f has a return type of String but is returning an int:

String f() => 3;

Common fixes

If the return type is correct, then replace the value being returned with a value of the correct type, possibly by converting the existing value:

String f() => 3.toString();

If the value is correct, then change the return type to match:

int f() => 3;

return_of_invalid_type_from_closure

The return type ‘{0}’ isn’t a ‘{1}’, as required by the closure’s context.

Description

The analyzer produces this diagnostic when the static type of a returned expression isn’t assignable to the return type that the closure is required to have.

Examples

The following code produces this diagnostic because f is defined to be a function that returns a String, but the closure assigned to it returns an int:

String Function(String) f = (s) => 3;

Common fixes

If the return type is correct, then replace the returned value with a value of the correct type, possibly by converting the existing value:

String Function(String) f = (s) => 3.toString();

return_without_value

The return value is missing after ‘return’.

Description

The analyzer produces this diagnostic when it finds a return statement without an expression in a function that declares a return type.

Examples

The following code produces this diagnostic because the function f is expected to return an int, but no value is being returned:

int f() {
  return;
}

Common fixes

Add an expression that computes the value to be returned:

int f() {
  return 0;
}

sdk_version_async_exported_from_core

The class ‘{0}’ wasn’t exported from ‘dart:core’ until version 2.1, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when either the class Future or Stream is referenced in a library that doesn’t import dart:async in code that has an SDK constraint whose lower bound is less than 2.1.0. In earlier versions, these classes weren’t defined in dart:core, so the import was necessary.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.1.0:

environment:
  sdk: '>=2.0.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

void f(Future f) {}

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the classes to be referenced:

environment:
  sdk: '>=2.1.0 <2.4.0'

If you need to support older versions of the SDK, then import the dart:async library.

import 'dart:async';

void f(Future f) {}

sdk_version_as_expression_in_const_context

The use of an as expression in a constant expression wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an as expression inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using an as expression in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const num n = 3;
const int i = n as int;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the expression to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use an as expression, or change the code so that the as expression isn’t in a constant context:

num x = 3;
int y = x as int;

sdk_version_bool_operator_in_const_context

The use of the operator ‘{0}’ for ‘bool’ operands in a constant context wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when any use of the &, |, or ^ operators on the class bool inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using these operators in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const bool a = true;
const bool b = false;
const bool c = a & b;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operators to be used:

environment:
 sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use these operators, or change the code so that the expression isn’t in a constant context:

const bool a = true;
const bool b = false;
bool c = a & b;

sdk_version_eq_eq_operator_in_const_context

Using the operator ‘==’ for non-primitive types wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when the operator == is used on a non-primitive type inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using this operator in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

class C {}
const C a = null;
const C b = null;
const bool same = a == b;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the operator to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use the == operator, or change the code so that the expression isn’t in a constant context:

class C {}
const C a = null;
const C b = null;
bool same = a == b;

sdk_version_extension_methods

Extension methods weren’t supported until version 2.6.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an extension declaration or an extension override is found in code that has an SDK constraint whose lower bound is less than 2.6.0. Using extensions wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.6.0:

environment:
 sdk: '>=2.4.0 <2.7.0'

In the package that has that pubspec, code like the following produces this diagnostic:

extension E on String {
  void sayHello() {
    print('Hello $this');
  }
}

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.6.0 <2.7.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of extensions. The most common way to do this is to rewrite the members of the extension as top-level functions (or methods) that take the value that would have been bound to this as a parameter:

void sayHello(String s) {
  print('Hello $s');
}

sdk_version_is_expression_in_const_context

The use of an is expression in a constant context wasn’t supported until version 2.3.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an is expression inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.3.2. Using an is expression in a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.2:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const x = 4;
const y = x is int ? 0 : 1;

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the expression to be used:

environment:
  sdk: '>=2.3.2 <2.4.0'

If you need to support older versions of the SDK, then either rewrite the code to not use the is operator, or, if that isn’t possible, change the code so that the is expression isn’t in a constant context:

const x = 4;
var y = x is int ? 0 : 1;

sdk_version_set_literal

Set literals weren’t supported until version 2.2, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when a set literal is found in code that has an SDK constraint whose lower bound is less than 2.2.0. Set literals weren’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.2.0:

environment:
  sdk: '>=2.1.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

var s = <int>{};

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.2.0 <2.4.0'

If you do need to support older versions of the SDK, then replace the set literal with code that creates the set without the use of a literal:

var s = new Set<int>();

sdk_version_ui_as_code

The for, if, and spread elements weren’t supported until version 2.3.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when a for, if, or spread element is found in code that has an SDK constraint whose lower bound is less than 2.3.0. Using a for, if, or spread element wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.3.0:

environment:
  sdk: '>=2.2.0 <2.4.0'

In the package that has that pubspec, code like the following produces this diagnostic:

var digits = [for (int i = 0; i < 10; i++) i];

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.3.0 <2.4.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of those elements:

var digits = _initializeDigits();

List<int> _initializeDigits() {
  var digits = <int>[];
  for (int i = 0; i < 10; i++) {
    digits.add(i);
  }
  return digits;
}

sdk_version_ui_as_code_in_const_context

The if and spread elements weren’t supported in constant expressions until version 2.5.0, but this code is required to be able to run on earlier versions.

Description

The analyzer produces this diagnostic when an if or spread element inside a constant context is found in code that has an SDK constraint whose lower bound is less than 2.5.0. Using an if or spread element inside a constant context wasn’t supported in earlier versions, so this code won’t be able to run against earlier versions of the SDK.

Examples

Here’s an example of a pubspec that defines an SDK constraint with a lower bound of less than 2.5.0:

environment:
  sdk: '>=2.4.0 <2.6.0'

In the package that has that pubspec, code like the following produces this diagnostic:

const a = [1, 2];
const b = [...a];

Common fixes

If you don’t need to support older versions of the SDK, then you can increase the SDK constraint to allow the syntax to be used:

environment:
  sdk: '>=2.5.0 <2.6.0'

If you need to support older versions of the SDK, then rewrite the code to not make use of those elements:

const a = [1, 2];
const b = [1, 2];

If that isn’t possible, change the code so that the element isn’t in a constant context:

const a = [1, 2];
var b = [...a];

static_access_to_instance_member

Instance member ‘{0}’ can’t be accessed using static access.

Description

The analyzer produces this diagnostic when a class name is used to access an instance field. Instance fields don’t exist on a class; they exist only on an instance of the class.

Examples

The following code produces this diagnostic because x is an instance field:

class C {
  static int a;

  int b;
}

int f() => C.b;

Common fixes

If you intend to access a static field, then change the name of the field to an existing static field:

class C {
  static int a;

  int b;
}

int f() => C.a;

If you intend to access the instance field, then use an instance of the class to access the field:

class C {
  static int a;

  int b;
}

int f(C c) => c.b;

subtype_of_disallowed_type

’‘{0}’ can’t be used as a superclass constraint.

Classes and mixins can’t implement ‘{0}’.

Classes can’t extend ‘{0}’.

Classes can’t mixin ‘{0}’.

Description

The analyzer produces this diagnostic when one of the restricted classes is used in either an extends, implements, with, or on clause. The classes bool, double, FutureOr, int, Null, num, and String are all restricted in this way, to allow for more efficient implementations.

Example

The following code produces this diagnostic because String is used in an extends clause:

class A extends String {}

The following code produces this diagnostic because String is used in an implements clause:

class B implements String {}

The following code produces this diagnostic because String is used in a with clause:

class C with String {}

The following code produces this diagnostic because String is used in an on clause:

mixin M on String {}

Common fixes

If a different type should be specified, then replace the type:

class A extends Object {}

If there isn’t a different type that would be appropriate, then remove the type, and possibly the whole clause:

class B {}

super_in_extension

The ‘super’ keyword can’t be used in an extension because an extension doesn’t have a superclass.

Description

The analyzer produces this diagnostic when a member declared inside an extension uses the super keyword . Extensions aren’t classes and don’t have superclasses, so the super keyword serves no purpose.

Examples

The following code produces this diagnostic because super can’t be used in an extension:

extension E on Object {
  String get displayString => super.toString();
}

Common fixes

Remove the super keyword :

extension E on Object {
  String get displayString => toString();
}

super_in_invalid_context

Invalid context for ‘super’ invocation.

Description

The analyzer produces this diagnostic when the keyword super is used outside of a instance method.

Examples

The following code produces this diagnostic because super is used in a top-level function:

void f() {
  super.f();
}

Common fixes

Rewrite the code to not use super.

switch_expression_not_assignable

Type ‘{0}’ of the switch expression isn’t assignable to the type ‘{1}’ of case expressions.

Description

The analyzer produces this diagnostic when the type of the expression in a switch statement isn’t assignable to the type of the expressions in the case clauses.

Example

The following code produces this diagnostic because the type of s (String) isn’t assignable to the type of 0 (int):

void f(String s) {
  switch (s) {
    case 0:
      break;
  }
}

Common fixes

If the type of the case expressions is correct, then change the expression in the switch statement to have the correct type:

void f(String s) {
  switch (int.parse(s)) {
    case 0:
      break;
  }
}

If the type of the switch expression is correct, then change the case expressions to have the correct type:

void f(String s) {
  switch (s) {
    case '0':
      break;
  }
}

throw_of_invalid_type

The type ‘{0}’ of the thrown expression must be assignable to ‘Object’.

Description

The analyzer produces this diagnostic when the type of the expression in a throw expression isn’t assignable to Object. It isn’t valid to throw null, so it isn’t valid to use an expression that might evaluate to null.

Example

The following code produces this diagnostic because s might be null:

void f(String? s) {
  throw s;
}

Common fixes

Add an explicit null check to the expression:

void f(String? s) {
  throw s!;
}

type_argument_not_matching_bounds

‘{0}’ doesn’t extend ‘{1}’.

Description

The analyzer produces this diagnostic when a type argument isn’t the same as or a subclass of the bounds of the corresponding type parameter.

Examples

The following code produces this diagnostic because String isn’t a subclass of num:

class A<E extends num> {}

var a = A<String>();

Common fixes

Change the type argument to be a subclass of the bounds:

class A<E extends num> {}

var a = A<int>();

type_test_with_undefined_name

The name ‘{0}’ isn’t defined, so it can’t be used in an ‘is’ expression.

Description

The analyzer produces this diagnostic when the name following the is in a type test expression isn’t defined.

Examples

The following code produces this diagnostic because the name Srting isn’t defined:

void f(Object o) {
  if (o is Srting) {
    // ...
  }
}

Common fixes

Replace the name with the name of a type:

void f(Object o) {
  if (o is String) {
    // ...
  }
}

unchecked_use_of_nullable_value

An expression whose value can be ‘null’ must be null-checked before it can be dereferenced.

Description

The analyzer produces this diagnostic when an expression whose type is potentially non-nullable is dereferenced without first verifying that the value isn’t null.

Example

The following code produces this diagnostic because s can be null at the point where it’s referenced:

void f(String? s) {
  if (s.length > 3) {
    // ...
  }
}

Common fixes

If the value really can be null, then add a test to ensure that members are only accessed when the value isn’t null:

void f(String? s) {
  if (s != null && s.length > 3) {
    // ...
  }
}

If the expression is a variable and the value should never be null, then change the type of the variable to be non-nullable:

void f(String s) {
  if (s.length > 3) {
    // ...
  }
}

If you believe that the value of the expression should never be null, but you can’t change the type of the variable, and you’re willing to risk having an exception thrown at runtime if you’re wrong, then you can assert that the value isn’t null:

void f(String? s) {
  if (s!.length > 3) {
    // ...
  }
}

undefined_annotation

Undefined name ‘{0}’ used as an annotation.

Description

The analyzer produces this diagnostic when a name that isn’t defined is used as an annotation.

Examples

The following code produces this diagnostic because the name undefined isn’t defined:

@undefined
void f() {}

Common fixes

If the name is correct, but it isn’t declared yet, then declare the name as a constant value:

const undefined = 'undefined';

@undefined
void f() {}

If the name is wrong, replace the name with the name of a valid constant:

@deprecated
void f() {}

Otherwise, remove the annotation.

undefined_class

Undefined class ‘{0}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a class but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because Piont isn’t defined:

class Point {}

void f(Piont p) {}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a class that is defined. The example above can be corrected by fixing the spelling of the class:

class Point {}

void f(Point p) {}

If the class is defined but isn’t visible, then you probably need to add an import.

undefined_constructor_in_initializer

The class ‘{0}’ doesn’t have a constructor named ‘{1}’.

The class ‘{0}’ doesn’t have an unnamed constructor.

Description

The analyzer produces this diagnostic when a superclass constructor is invoked in the initializer list of a constructor, but the superclass doesn’t define the constructor being invoked.

Examples

The following code produces this diagnostic because A doesn’t have an unnamed constructor:

class A {
  A.n();
}
class B extends A {
  B() : super();
}

The following code produces this diagnostic because A doesn’t have a constructor named m:

class A {
  A.n();
}
class B extends A {
  B() : super.m();
}

Common fixes

If the superclass defines a constructor that should be invoked, then change the constructor being invoked:

class A {
  A.n();
}
class B extends A {
  B() : super.n();
}

If the superclass doesn’t define an appropriate constructor, then define the constructor being invoked:

class A {
  A.m();
  A.n();
}
class B extends A {
  B() : super.m();
}

undefined_enum_constant

There’s no constant named ‘{0}’ in ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of an enum constant, and the name either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because E doesn’t define a constant named c:

enum E {a, b}

var e = E.c;

Common fixes

If the constant should be defined, then add it to the declaration of the enum:

enum E {a, b, c}

var e = E.c;

If the constant shouldn’t be defined, then change the name to the name of an existing constant:

enum E {a, b}

var e = E.b;

undefined_extension_getter

The getter ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a getter, but the getter isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static getter is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance getter named b:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').b;
}

The following code produces this diagnostic because the extension E doesn’t declare a static getter named a:

extension E on String {}

var x = E.a;

Common fixes

If the name of the getter is incorrect, then change it to the name of an existing getter:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').a;
}

If the name of the getter is correct but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  String get a => 'a';
}

extension F on String {
  String get b => 'b';
}

void f() {
  F('c').b;
}

If the name of the getter and extension are both correct, but the getter isn’t defined, then define the getter:

extension E on String {
  String get a => 'a';
  String get b => 'z';
}

extension F on String {
  String get b => 'b';
}

void f() {
  E('c').b;
}

undefined_extension_method

The method ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a method, but the method isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static method is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance method named b:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').b();
}

The following code produces this diagnostic because the extension E doesn’t declare a static method named a:

extension E on String {}

var x = E.a();

Common fixes

If the name of the method is incorrect, then change it to the name of an existing method:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').a();
}

If the name of the method is correct, but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  String a() => 'a';
}

extension F on String {
  String b() => 'b';
}

void f() {
  F('c').b();
}

If the name of the method and extension are both correct, but the method isn’t defined, then define the method:

extension E on String {
  String a() => 'a';
  String b() => 'z';
}

extension F on String {
  String b() => 'b';
}

void f() {
  E('c').b();
}

undefined_extension_setter

The setter ‘{0}’ isn’t defined for the extension ‘{1}’.

Description

The analyzer produces this diagnostic when an extension override is used to invoke a setter, but the setter isn’t defined by the specified extension. The analyzer also produces this diagnostic when a static setter is referenced but isn’t defined by the specified extension.

Examples

The following code produces this diagnostic because the extension E doesn’t declare an instance setter named b:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').b = 'd';
}

The following code produces this diagnostic because the extension E doesn’t declare a static setter named a:

extension E on String {}

void f() {
  E.a = 3;
}

Common fixes

If the name of the setter is incorrect, then change it to the name of an existing setter:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').a = 'd';
}

If the name of the setter is correct, but the name of the extension is wrong, then change the name of the extension to the correct name:

extension E on String {
  set a(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  F('c').b = 'd';
}

If the name of the setter and extension are both correct, but the setter isn’t defined, then define the setter:

extension E on String {
  set a(String v) {}
  set b(String v) {}
}

extension F on String {
  set b(String v) {}
}

void f() {
  E('c').b = 'd';
}

undefined_function

The function ‘{0}’ isn’t defined.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a function but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because the name emty isn’t defined:

List<int> empty() => [];

void main() {
  print(emty());
}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a function that is defined. The example above can be corrected by fixing the spelling of the function:

List<int> empty() => [];

void main() {
  print(empty());
}

If the function is defined but isn’t visible, then you probably need to add an import or re-arrange your code to make the function visible.

undefined_getter

The getter ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a getter but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because String has no member named len:

int f(String s) => s.len;

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a getter that is defined. The example above can be corrected by fixing the spelling of the getter:

int f(String s) => s.length;

undefined_hidden_name

The library ‘{0}’ doesn’t export a member with the hidden name ‘{1}’.

Description

The analyzer produces this diagnostic when a hide combinator includes a name that isn’t defined by the library being imported.

Examples

The following code produces this diagnostic because dart:math doesn’t define the name String:

import 'dart:math' hide String, max;

var x = min(0, 1);

Common fixes

If a different name should be hidden, then correct the name. Otherwise, remove the name from the list:

import 'dart:math' hide max;

var x = min(0, 1);

undefined_identifier

Undefined name ‘{0}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because the name rihgt isn’t defined:

int min(int left, int right) => left <= rihgt ? left : right;

Common fixes

If the identifier isn’t defined, then either define it or replace it with an identifier that is defined. The example above can be corrected by fixing the spelling of the variable:

int min(int left, int right) => left <= right ? left : right;

If the identifier is defined but isn’t visible, then you probably need to add an import or re-arrange your code to make the identifier visible.

undefined_identifier_await

Undefined name ‘await’ in function body not marked with ‘async’.

Description

The analyzer produces this diagnostic when the name await is used in a method or function body without being declared, and the body isn’t marked with the async keyword. The name await only introduces an await expression in an asynchronous function.

Example

The following code produces this diagnostic because the name await is used in the body of f even though the body of f isn’t marked with the async keyword:

void f(p) { await p; }

Common fixes

Add the keyword async to the function body:

void f(p) async { await p; }

undefined_method

The method ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a method but either isn’t defined or isn’t visible in the scope in which it’s being referenced.

Examples

The following code produces this diagnostic because the identifier removeMiddle isn’t defined:

int f(List<int> l) => l.removeMiddle();

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a method that is defined. The example above can be corrected by fixing the spelling of the method:

int f(List<int> l) => l.removeLast();

undefined_named_parameter

The named parameter ‘{0}’ isn’t defined.

Description

The analyzer produces this diagnostic when a method or function invocation has a named argument, but the method or function being invoked doesn’t define a parameter with the same name.

Examples

The following code produces this diagnostic because m doesn’t declare a named parameter named a:

class C {
  m({int b}) {}
}

void f(C c) {
  c.m(a: 1);
}

Common fixes

If the argument name is mistyped, then replace it with the correct name. The example above can be fixed by changing a to b:

class C {
  m({int b}) {}
}

void f(C c) {
  c.m(b: 1);
}

If a subclass adds a parameter with the name in question, then cast the receiver to the subclass:

class C {
  m({int b}) {}
}

class D extends C {
  m({int a, int b}) {}
}

void f(C c) {
  (c as D).m(a: 1);
}

If the parameter should be added to the function, then add it:

class C {
  m({int a, int b}) {}
}

void f(C c) {
  c.m(a: 1);
}

undefined_operator

The operator ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when a user-definable operator is invoked on an object for which the operator isn’t defined.

Examples

The following code produces this diagnostic because the class C doesn’t define the operator +:

class C {}

C f(C c) => c + 2;

Common fixes

If the operator should be defined for the class, then define it:

class C {
  C operator +(int i) => this;
}

C f(C c) => c + 2;

undefined_prefixed_name

The name ‘{0}’ is being referenced through the prefix ‘{1}’, but it isn’t defined in any of the libraries imported using that prefix.

Description

The analyzer produces this diagnostic when a prefixed identifier is found where the prefix is valid, but the identifier isn’t declared in any of the libraries imported using that prefix.

Examples

The following code produces this diagnostic because dart:core doesn’t define anything named a:

import 'dart:core' as p;

void f() {
  p.a;
}

Common fixes

If the library in which the name is declared isn’t imported yet, add an import for the library.

If the name is wrong, then change it to one of the names that’s declared in the imported libraries.

undefined_setter

The setter ‘{0}’ isn’t defined for the type ‘{1}’.

Description

The analyzer produces this diagnostic when it encounters an identifier that appears to be the name of a setter but either isn’t defined or isn’t visible in the scope in which the identifier is being referenced.

Examples

The following code produces this diagnostic because there isn’t a setter named z:

class C {
  int x = 0;
  void m(int y) {
    this.z = y;
  }
}

Common fixes

If the identifier isn’t defined, then either define it or replace it with the name of a setter that is defined. The example above can be corrected by fixing the spelling of the setter:

class C {
  int x = 0;
  void m(int y) {
    this.x = y;
  }
}

undefined_shown_name

The library ‘{0}’ doesn’t export a member with the shown name ‘{1}’.

Description

The analyzer produces this diagnostic when a show combinator includes a name that isn’t defined by the library being imported.

Examples

The following code produces this diagnostic because dart:math doesn’t define the name String:

import 'dart:math' show min, String;

var x = min(0, 1);

Common fixes

If a different name should be shown, then correct the name. Otherwise, remove the name from the list:

import 'dart:math' show min;

var x = min(0, 1);

undefined_super_member

The getter ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The method ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The operator ‘{0}’ isn’t defined in a superclass of ‘{1}’.

The setter ‘{0}’ isn’t defined in a superclass of ‘{1}’.

Description

The analyzer produces this diagnostic when an inherited member (method, getter, setter, or operator) is referenced using super, but there’s no member with that name in the superclass chain.

Examples

The following code produces this diagnostic because Object doesn’t define a method named n:

class C {
  void m() {
    super.n();
  }
}

The following code produces this diagnostic because Object doesn’t define a getter named g:

class C {
  void m() {
    super.g;
  }
}

Common fixes

If the inherited member you intend to invoke has a different name, then make the name of the invoked member match the inherited member.

If the member you intend to invoke is defined in the same class, then remove the super..

If the member isn’t defined, then either add the member to one of the superclasses or remove the invocation.

unnecessary_cast

Unnecessary cast.

Description

The analyzer produces this diagnostic when the value being cast is already known to be of the type that it’s being cast to.

Examples

The following code produces this diagnostic because n is already known to be an int as a result of the is test:

void f(num n) {
  if (n is int) {
    (n as int).isEven;
  }
}

Common fixes

Remove the unnecessary cast:

void f(num n) {
  if (n is int) {
    n.isEven;
  }
}

unnecessary_non_null_assertion

The ‘!’ will have no effect because the receiver can’t be null.

Description

The analyzer produces this diagnostic when the operand of the ! operator can’t be null.

Example

The following code produces this diagnostic because x can’t be null:

int f(int x) {
  return x!;
}

Common fixes

Remove the null check operator (!):

int f(int x) {
  return x;
}

unnecessary_null_comparison

The operand can’t be null, so the condition is always false.

The operand can’t be null, so the condition is always true.

Description

The analyzer produces this diagnostic when it finds an equality comparison (either == or !=) with one operand of null and the other operand can’t be null. Such comparisons are always either true or false, so they serve no purpose.

Example

The following code produces this diagnostic because x can never be null, so the comparison always evaluates to true:

void f(int x) {
  if (x != null) {
    print(x);
  }
}

The following code produces this diagnostic because x can never be null, so the comparison always evaluates to false:

void f(int x) {
  if (x == null) {
    throw ArgumentError("x can't be null");
  }
}

Common fixes

If the other operand should be able to be null, then change the type of the operand:

void f(int? x) {
  if (x != null) {
    print(x);
  }
}

If the other operand really can’t be null, then remove the condition:

void f(int x) {
  print(x);
}

unqualified_reference_to_static_member_of_extended_type

Static members from the extended type or one of its superclasses must be qualified by the name of the defining type.

Description

The analyzer produces this diagnostic when an undefined name is found, and the name is the same as a static member of the extended type or one of its superclasses.

Examples

The following code produces this diagnostic because m is a static member of the extended type C:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    m();
  }
}

Common fixes

If you’re trying to reference a static member that’s declared outside the extension, then add the name of the class or extension before the reference to the member:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    C.m();
  }
}

If you’re referencing a member that isn’t declared yet, add a declaration:

class C {
  static void m() {}
}

extension E on C {
  void f() {
    m();
  }

  void m() {}
}

unused_catch_clause

The exception variable ‘{0}’ isn’t used, so the ‘catch’ clause can be removed.

Description

The analyzer produces this diagnostic when a catch clause is found, and neither the exception parameter nor the optional stack trace parameter are used in the catch block.

Examples

The following code produces this diagnostic because e isn’t referenced:

void f() {
  try {
    int.parse(';');
  } on FormatException catch (e) {
    // ignored
  }
}

Common fixes

Remove the unused catch clause:

void f() {
  try {
    int.parse(';');
  } on FormatException {
    // ignored
  }
}

unused_catch_stack

The stack trace variable ‘{0}’ isn’t used and can be removed.

Description

The analyzer produces this diagnostic when the stack trace parameter in a catch clause isn’t referenced within the body of the catch block.

Examples

The following code produces this diagnostic because stackTrace isn’t referenced:

void f() {
  try {
    // ...
  } catch (exception, stackTrace) {
    // ...
  }
}

Common fixes

If you need to reference the stack trace parameter, then add a reference to it. Otherwise, remove it:

void f() {
  try {
    // ...
  } catch (exception) {
    // ...
  }
}

unused_element

A value for optional parameter ‘{0}’ isn’t ever given.

The declaration ‘{0}’ isn’t referenced.

Description

The analyzer produces this diagnostic when a private declaration isn’t referenced in the library that contains the declaration. The following kinds of declarations are analyzed:

  • Private top-level declarations, such as classes, enums, mixins, typedefs, top-level variables, and top-level functions
  • Private static and instance methods
  • Optional parameters of private functions for which a value is never passed, even when the parameter doesn’t have a private name

Examples

Assuming that no code in the library references _C, the following code produces this diagnostic:

class _C {}

Assuming that no code in the library passes a value for y in any invocation of _m, the following code produces this diagnostic:

class C {
  void _m(int x, [int y]) {}

  void n() => _m(0);
}

Common fixes

If the declaration isn’t needed, then remove it:

class C {
  void _m(int x) {}

  void n() => _m(0);
}

If the declaration is intended to be used, then add the code to use it.

unused_field

The value of the field ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a private field is declared but never read, even if it’s written in one or more places.

Examples

The following code produces this diagnostic because _x isn’t referenced anywhere in the library:

class Point {
  int _x;
}

Common fixes

If the field isn’t needed, then remove it.

If the field was intended to be used, then add the missing code.

unused_import

Unused import: ‘{0}’.

Description

The analyzer produces this diagnostic when an import isn’t needed because none of the names that are imported are referenced within the importing library.

Examples

The following code produces this diagnostic because nothing defined in dart:async is referenced in the library:

import 'dart:async';

void main() {}

Common fixes

If the import isn’t needed, then remove it.

If some of the imported names are intended to be used, then add the missing code.

unused_label

The label ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a label that isn’t used is found.

Examples

The following code produces this diagnostic because the label loop isn’t referenced anywhere in the method:

void f(int limit) {
  loop: for (int i = 0; i < limit; i++) {
    print(i);
  }
}

Common fixes

If the label isn’t needed, then remove it:

void f(int limit) {
  for (int i = 0; i < limit; i++) {
    print(i);
  }
}

If the label is needed, then use it:

void f(int limit) {
  loop: for (int i = 0; i < limit; i++) {
    print(i);
    break loop;
  }
}

unused_local_variable

The value of the local variable ‘{0}’ isn’t used.

Description

The analyzer produces this diagnostic when a local variable is declared but never read, even if it’s written in one or more places.

Examples

The following code produces this diagnostic because the value of count is never read:

void main() {
  int count = 0;
}

Common fixes

If the variable isn’t needed, then remove it.

If the variable was intended to be used, then add the missing code.

unused_shown_name

The name {0} is shown, but isn’t used.

Description

The analyzer produces this diagnostic when a show combinator includes a name that isn’t used within the library. Because it isn’t referenced, the name can be removed.

Examples

The following code produces this diagnostic because the function max isn’t used:

import 'dart:math' show min, max;

var x = min(0, 1);

Common fixes

Either use the name or remove it:

import 'dart:math' show min;

var x = min(0, 1);

uri_does_not_exist

Target of URI doesn’t exist: ‘{0}’.

Description

The analyzer produces this diagnostic when an import, export, or part directive is found where the URI refers to a file that doesn’t exist.

Examples

If the file lib.dart doesn’t exist, the following code produces this diagnostic:

import 'lib.dart';

Common fixes

If the URI was mistyped or invalid, then correct the URI.

If the URI is correct, then create the file.

uri_has_not_been_generated

Target of URI hasn’t been generated: ‘{0}’.

Description

The analyzer produces this diagnostic when an import, export, or part directive is found where the URI refers to a file that doesn’t exist and the name of the file ends with a pattern that’s commonly produced by code generators, such as one of the following:

  • .g.dart
  • .pb.dart
  • .pbenum.dart
  • .pbserver.dart
  • .pbjson.dart
  • .template.dart

Examples

If the file lib.g.dart doesn’t exist, the following code produces this diagnostic:

import 'lib.g.dart';

Common fixes

If the file is a generated file, then run the generator that generates the file.

If the file isn’t a generated file, then check the spelling of the URI or create the file.

use_of_void_result

This expression has a type of ‘void’ so its value can’t be used.

Description

The analyzer produces this diagnostic when it finds an expression whose type is void, and the expression is used in a place where a value is expected, such as before a member access or on the right-hand side of an assignment.

Examples

The following code produces this diagnostic because f doesn’t produce an object on which toString can be invoked:

void f() {}

void g() {
  f().toString();
}

Common fixes

Either rewrite the code so that the expression has a value or rewrite the code so that it doesn’t depend on the value.

variable_type_mismatch

A value of type ‘{0}’ can’t be assigned to a const variable of type ‘{1}’.

Description

The analyzer produces this diagnostic when the evaluation of a constant expression would result in a CastException.

Examples

The following code produces this diagnostic because the value of x is an int, which can’t be assigned to y because an int isn’t a String:

const Object x = 0;
const String y = x;

Common fixes

If the declaration of the constant is correct, then change the value being assigned to be of the correct type:

const Object x = 0;
const String y = '$x';

If the assigned value is correct, then change the declaration to have the correct type:

const Object x = 0;
const int y = x;

wrong_number_of_parameters_for_operator

Operator ‘{0}’ should declare exactly {1} parameters, but {2} found.

Description

The analyzer produces this diagnostic when a declaration of an operator has the wrong number of parameters.

Examples

The following code produces this diagnostic because the operator + must have a single parameter corresponding to the right operand:

class C {
  int operator +(a, b) => 0;
}

Common fixes

Add or remove parameters to match the required number:

class C {
  int operator +(a) => 0;
}

wrong_number_of_parameters_for_setter

Setters must declare exactly one required positional parameter.

Description

The analyzer produces this diagnostic when a setter is found that doesn’t declare exactly one required positional parameter.

Examples

The following code produces this diagnostic because the setter s declares two required parameters:

class C {
  set s(int x, int y) {}
}

The following code produces this diagnostic because the setter s declares one optional parameter:

class C {
  set s([int x]) {}
}

Common fixes

Change the declaration so that there’s exactly one required positional parameter:

class C {
  set s(int x) {}
}

wrong_number_of_type_arguments

The type ‘{0}’ is declared with {1} type parameters, but {2} type arguments were given.

Description

The analyzer produces this diagnostic when a type that has type parameters is used and type arguments are provided, but the number of type arguments isn’t the same as the number of type parameters.

The analyzer also produces this diagnostic when a constructor is invoked and the number of type arguments doesn’t match the number of type parameters declared for the class.

Examples

The following code produces this diagnostic because C has one type parameter but two type arguments are provided when it is used as a type annotation:

class C<E> {}

void f(C<int, int> x) {}

The following code produces this diagnostic because C declares one type parameter, but two type arguments are provided when creating an instance:

class C<E> {}

var c = C<int, int>();

Common fixes

Add or remove type arguments, as necessary, to match the number of type parameters defined for the type:

class C<E> {}

void f(C<int> x) {}

wrong_number_of_type_arguments_method

The method ‘{0}’ is declared with {1} type parameters, but {2} type arguments are given.

Description

The analyzer produces this diagnostic when a method or function is invoked with a different number of type arguments than the number of type parameters specified in its declaration. There must either be no type arguments or the number of arguments must match the number of parameters.

Example

The following code produces this diagnostic because the invocation of the method m has two type arguments, but the declaration of m only has one type parameter:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m<int, int>(2);

Common fixes

If the type arguments are necessary, then make them match the number of type parameters by either adding or removing type arguments:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m<int>(2);

If the type arguments aren’t necessary, then remove them:

class C {
  int m<A>(A a) => 0;
}

int f(C c) => c.m(2);

undefined_super_method

See undefined_super_member.