Fixing type promotion failures

Type promotion occurs when flow analysis can soundly confirm a variable with a nullable type is not null, and that it will not change from that point on. Many circumstances can weaken a type's soundness, causing type promotion to fail.

This page lists reasons why type promotion failures occur, with tips on how to fix them. To learn more about flow analysis and type promotion, check out the Understanding null safety page.

The cause: You're trying to promote a field, but field promotion is language versioned, and your code is set to a language version before 3.2.

If you're already using an SDK version >= Dart 3.2, your code might still be explicitly targeted for an earlier language version. This can happen either because:

• Your pubspec.yaml declares an SDK constraint with a lower bound below 3.2, or
• You have a // @dart=version comment at the top of the file, where version is lower than 3.2.

Example:

// @dart=3.1

class C {
  final int? _i;
  C(this._i);

  void f() {
    if (_i != null) {
      int i = _i;  // ERROR
    }
  }
}

Message:

'_i' refers to a field. It couldn't be promoted because field promotion is only available in Dart 3.2 and above.

Solution:

Ensure your library isn't using a language version earlier than 3.2. Check the top of your file for an outdated // @dart=version comment, or your pubspec.yaml for an outdated SDK constraint lower-bound.

The cause: You're trying to promote a property, but only local variables can be promoted in Dart versions earlier than 3.2, and you are using a version earlier than 3.2.

Example:

class C {
  int? i;
  void f() {
    if (i == null) return;
    print(i.isEven);       // ERROR
  }
}

Message:

'i' refers to a property so it couldn't be promoted.

Solution:

If you are using Dart 3.1 or earlier, upgrade to 3.2 or later.

If you need to keep using an older version, read Other causes and workarounds

The remaining examples on this page document reasons for promotion failures unrelated to version inconsistencies, for both field and local variable failures, with examples and workarounds.

In general, the usual fixes for promotion failures are one or more of the following:

• Assign the property's value to a local variable with the non-nullable type you need.
• Add an explicit null check (for example, i == null).
• Use ! or as as a redundant check if you're sure an expression can't be null.

Here's an example of creating a local variable (which can be named i) that holds the value of i:

class C {
  int? i;
  void f() {
    final i = this.i;
    if (i == null) return;
    print(i.isEven);
  }
}

This example features an instance field, but it could instead use an instance getter, a static field or getter, a top-level variable or getter, or this.

And here's an example of using i!:

print(i!.isEven);

The cause: You're trying to promote this, but type promotion for this is not yet supported.

One common this promotion scenario is when writing extension methods. If the on type of the extension method is a nullable type, you'd want to do a null check to see whether this is null:

Example:

extension on int? {
  int get valueOrZero {
    return this == null ? 0 : this; // ERROR
  }
}

Message:

`this` can't be promoted.

Solution:

Create a local variable to hold the value of this, then perform the null check.

extension on int? {
  int get valueOrZero {
    final self = this;
    return self == null ? 0 : self;
  }
}

The cause: You're trying to promote a field, but the field is not private.

It's possible for other libraries in your program to override public fields with a getter. Because getters might not return a stable value, and the compiler can't know what other libraries are doing, non-private fields cannot be promoted.

Example:

class Example {
  final int? value;
  Example(this.value);
}

void test(Example x) {
  if (x.value != null) {
    print(x.value + 1); // ERROR
  }
}

Message:

'value' refers to a public property so it couldn't be promoted.

Solution:

Making the field private lets the compiler be sure that no outside libraries could possibly override its value, so it's safe to promote.

class Example {
  final int? _value;
  Example(this._value);
}

void test(Example x) {
  if (x._value != null) {
    print(x._value + 1);
  }
}

The cause: You're trying to promote a field, but the field is not final.

To the compiler, non-final fields could, in principle, be modified any time between the time they're tested and the time they're used. So it's not safe for the compiler to promote a non-final nullable type to a non-nullable type.

Example:

class Example {
  int? _mutablePrivateField;
  Example(this._mutablePrivateField);

  void f() {
    if (_mutablePrivateField != null) {
      int i = _mutablePrivateField; // ERROR
    }
  }
}

Message:

'_mutablePrivateField' refers to a non-final field so it couldn't be promoted.

Solution:

Make the field final:

class Example {
  final int? _immutablePrivateField;
  Example(this._immutablePrivateField);

  void f() {
    if (_immutablePrivateField != null) {
      int i = _immutablePrivateField; // OK
    }
  }
}

The cause: You're trying to promote a getter, but only instance fields can be promoted, not instance getters.

The compiler has no way to guarantee that a getter returns the same result every time. Because their stability can't be confirmed, getters are not safe to promote.

Example:

import 'dart:math';

abstract class Example {
  int? get _value => Random().nextBool() ? 123 : null;
}

void f(Example x) {
  if (x._value != null) {
    print(x._value.isEven); // ERROR
  }
}

Message:

'_value' refers to a getter so it couldn't be promoted.

Solution:

Assign the getter to a local variable:

import 'dart:math';

abstract class Example {
  int? get _value => Random().nextBool() ? 123 : null;
}

void f(Example x) {
  final value = x._value;
  if (value != null) {
    print(value.isEven); // OK
  }
}

The cause: You're trying to promote a field, but the field is marked external.

External fields don't promote because they are essentially external getters; their implementation is code from outside of Dart, so there's no guarantee for the compiler that an external field will return the same value each time it's called.

Example:

class Example {
  external final int? _externalField;

  void f() {
    if (_externalField != null) {
      print(_externalField.isEven); // ERROR
    }
  }
}

Message:

'_externalField' refers to an external field so it couldn't be promoted.

Solution:

Assign the external field's value to a local variable:

class Example {
  external final int? _externalField;

  void f() {
    final i = _externalField;
    if (i != null) {
      print(i.isEven); // OK
    }
  }
}

The cause: You're trying to promote a field, but another class in the same library contains a concrete getter with the same name.

Example:

import 'dart:math';

class Example {
  final int? _overridden;
  Example(this._overridden);
}

class Override implements Example {
  @override
  int? get _overridden => Random().nextBool() ? 1 : null;
}

void testParity(Example x) {
  if (x._overridden != null) {
    print(x._overridden.isEven); // ERROR
  }
}

Message:

'_overriden' couldn't be promoted because there is a conflicting getter in class 'Override'.

Solution:

If the getter and field are related and need to share their name (like when one of them overrides the other, as in the example above), then you can enable type promotion by assigning the value to a local variable:

import 'dart:math';

class Example {
  final int? _overridden;
  Example(this._overridden);
}

class Override implements Example {
  @override
  int? get _overridden => Random().nextBool() ? 1 : null;
}

void testParity(Example x) {
  final i = x._overridden;
  if (i != null) {
    print(i.isEven); // OK
  }
}

Note that in the above example it's clear why it's unsafe to promote the field _overridden: because there's an override relationship between the field and the getter. However, a conflicting getter will prevent field promotion even if the classes are unrelated. For example:

import 'dart:math';

class Example {
  final int? _i;
  Example(this._i);
}

class Unrelated {
  int? get _i => Random().nextBool() ? 1 : null;
}

void f(Example x) {
  if (x._i != null) {
    int i = x._i; // ERROR
  }
}

Another library might contain a class that combines the two unrelated classes together into the same class hierarchy, which would cause the reference in function f to x._i to get dispatched to Unrelated._i. For example:

class Surprise extends Unrelated implements Example {}

void main() {
  f(Surprise());
}

Solution:

If the field and the conflicting entity are truly unrelated, you can work around the problem by giving them different names:

class Example {
  final int? _i;
  Example(this._i);
}

class Unrelated {
  int? get _j => Random().nextBool() ? 1 : null;
}

void f(Example x) {
  if (x._i != null) {
    int i = x._i; // OK
  }
}

The cause: You're trying to promote a field, but another class in the same library contains a field with the same name that isn't promotable (for any of the other reasons listed on this page).

Example:

class Example {
  final int? _overridden;
  Example(this._overridden);
}

class Override implements Example {
  @override
  int? _overridden;
}

void f(Example x) {
  if (x._overridden != null) {
    print(x._overridden.isEven); // ERROR
  }
}

This example fails because at runtime, x might actually be an instance of Override, so promotion would not be sound.

Message:

'overridden' couldn't be promoted because there is a conflicting non-promotable field in class 'Override'.

Solution:

If the fields are actually related and need to share a name, then you can enable type promotion by assigning the value to a final local variable to promote:

class Example {
  final int? _overridden;
  Example(this._overridden);
}

class Override implements Example {
  @override
  int? _overridden;
}

void f(Example x) {
  final i = x._overridden;
  if (i != null) {
    print(i.isEven); // OK
  }
}

If the fields are unrelated, then rename one of the fields, so they don't conflict. Read the Note about unrelated classes.

The cause: You're trying to promote a field that is private and final, but another class in the same library contains an implicit noSuchMethod forwarder with the same name as the field.

This is unsound because there's no guarantee that noSuchMethod will return a stable value from one invocation to the next.

Example:

import 'package:mockito/mockito.dart';

class Example {
  final int? _i;
  Example(this._i);
}

class MockExample extends Mock implements Example {}

void f(Example x) {
  if (x._i != null) {
    int i = x._i; // ERROR
  }
}

In this example, _i can't be promoted because it could resolve to the unsound implicit noSuchMethod forwarder (also named _i) that the compiler generates inside MockExample.

The compiler creates this implicit implementation of _i because MockExample promises to support a getter for _i when it implements Example in its declaration, but doesn't fulfill that promise. So, the undefined getter implementation is handled by Mock's noSuchMethod definition, which creates an implicit noSuchMethod forwarder of the same name.

The failure can also occur between fields in unrelated classes.

Message:

'_i' couldn't be promoted because there is a conflicting noSuchMethod forwarder in class 'MockExample'.

Solution:

Define the getter in question so that noSuchMethod doesn't have to implicitly handle its implementation:

import 'package:mockito/mockito.dart';

class Example {
  final int? _i;
  Example(this._i);
}

class MockExample extends Mock implements Example {
  @override
  late final int? _i;
}

void f(Example x) {
  if (x._i != null) {
    int i = x._i; // OK
  }
}

The getter is declared late to be consistent with how mocks are generally used; it's not necessary to declare the getter late to solve this type promotion failure in scenarios not involving mocks.

The cause: You're trying to promote a variable that might have been written to since it was promoted.

Example:

void f(bool b, int? i, int? j) {
  if (i == null) return;
  if (b) {
    i = j;           // (1)
  }
  if (!b) {
    print(i.isEven); // (2) ERROR
  }
}

Solution:

In this example, when flow analysis hits (1), it demotes i from non-nullable int back to nullable int?. A human can tell that the access at (2) is safe because there's no code path that includes both (1) and (2), but flow analysis isn't smart enough to see that, because it doesn't track correlations between conditions in separate if statements.

You might fix the problem by combining the two if statements:

void f(bool b, int? i, int? j) {
  if (i == null) return;
  if (b) {
    i = j;
  } else {
    print(i.isEven);
  }
}

In straight-line control flow cases like these (no loops), flow analysis takes into account the right hand side of the assignment when deciding whether to demote. As a result, another way to fix this code is to change the type of j to int.

void f(bool b, int? i, int j) {
  if (i == null) return;
  if (b) {
    i = j;
  }
  if (!b) {
    print(i.isEven);
  }
}

The cause: You're trying to promote something that might have been written to in a previous iteration of a loop, and so the promotion was invalidated.

Example:

void f(Link? p) {
  if (p != null) return;
  while (true) {    // (1)
    print(p.value); // (2) ERROR
    var next = p.next;
    if (next == null) break;
    p = next;       // (3)
  }
}

When flow analysis reaches (1), it looks ahead and sees the write to p at (3). But because it's looking ahead, it hasn't yet figured out the type of the right-hand side of the assignment, so it doesn't know whether it's safe to retain the promotion. To be safe, it invalidates the promotion.

Solution:

You can fix this problem by moving the null check to the top of the loop:

void f(Link? p) {
  while (p != null) {
    print(p.value);
    p = p.next;
  }
}

This situation can also arise in switch statements if a case block has a label, because you can use labeled switch statements to construct loops:

void f(int i, int? j, int? k) {
  if (j == null) return;
  switch (i) {
    label:
    case 0:
      print(j.isEven); // ERROR
      j = k;
      continue label;
  }
}

Again, you can fix the problem by moving the null check to the top of the loop:

void f(int i, int? j, int? k) {
  switch (i) {
    label:
    case 0:
      if (j == null) return;
      print(j.isEven);
      j = k;
      continue label;
  }
}

The cause: The variable might have been written to in a try block, and execution is now in a catch block.

Example:

void f(int? i, int? j) {
  if (i == null) return;
  try {
    i = j;                 // (1)
    // ... Additional code ...
    if (i == null) return; // (2)
    // ... Additional code ...
  } catch (e) {
    print(i.isEven);       // (3) ERROR
  }
}

In this case, flow analysis doesn't consider i.isEven (3) safe, because it has no way of knowing when in the try block the exception might have occurred, so it conservatively assumes that it might have happened between (1) and (2), when i was potentially null.

Similar situations can occur between try and finally blocks, and between catch and finally blocks. Because of a historical artifact of how the implementation was done, these try/catch/finally situations don't take into account the right-hand side of the assignment, similar to what happens in loops.

Solution:

To fix the problem, make sure that the catch block doesn't rely on assumptions about the state of variables that get changed inside the try block. Remember, the exception might occur at any time during the try block, possibly when i is null.

The safest solution is to add a null check inside the catch block:

try {
  // ···
} catch (e) {
  if (i != null) {
    print(i.isEven); // (3) OK due to the null check in the line above.
  } else {
    // Handle the case where i is null.
  }
}

Or, if you're sure that an exception can't occur while i is null, just use the ! operator:

try {
  // ···
} catch (e) {
  print(i!.isEven); // (3) OK because of the `!`.
}

The cause: You're trying to promote to a type isn't a subtype of the variable's current promoted type (or wasn't a subtype at the time of the promotion attempt).

Example:

void f(Object o) {
  if (o is Comparable /* (1) */) {
    if (o is Pattern /* (2) */) {
      print(o.matchAsPrefix('foo')); // (3) ERROR
    }
  }
}

In this example, o is promoted to Comparable at (1), but it isn't promoted to Pattern at (2), because Pattern isn't a subtype of Comparable. (The rationale is that if it did promote, then you wouldn't be able to use methods on Comparable.) Note that just because Pattern isn't a subtype of Comparable doesn't mean the code at (3) is dead; o might have a type—like String—that implements both Comparable and Pattern.

Solution:

One possible solution is to create a new local variable so that the original variable is promoted to Comparable, and the new variable is promoted to Pattern:

void f(Object o) {
  if (o is Comparable /* (1) */) {
    Object o2 = o;
    if (o2 is Pattern /* (2) */) {
      print(
          o2.matchAsPrefix('foo')); // (3) OK; o2 was promoted to `Pattern`.
    }
  }
}

However, someone who edits the code later might be tempted to change Object o2 to var o2. That change gives o2 a type of Comparable, which brings back the problem of the object not being promotable to Pattern.

A redundant type check might be a better solution:

void f(Object o) {
  if (o is Comparable /* (1) */) {
    if (o is Pattern /* (2) */) {
      print((o as Pattern).matchAsPrefix('foo')); // (3) OK
    }
  }
}

Another solution that sometimes works is when you can use a more precise type. If line 3 cares only about strings, then you can use String in your type check. Because String is a subtype of Comparable, the promotion works:

void f(Object o) {
  if (o is Comparable /* (1) */) {
    if (o is String /* (2) */) {
      print(o.matchAsPrefix('foo')); // (3) OK
    }
  }
}

The cause: The variable has been write-captured by a local function or function expression.

Example:

void f(int? i, int? j) {
  var foo = () {
    i = j;
  };
  // ... Use foo ... 
  if (i == null) return; // (1)
  // ... Additional code ...
  print(i.isEven);       // (2) ERROR
}

Flow analysis reasons that as soon as the definition of foo is reached, it might get called at any time, therefore it's no longer safe to promote i at all. As with loops, this demotion happens regardless of the type of the right hand side of the assignment.

Solution:

Sometimes it's possible to restructure the logic so that the promotion is before the write capture:

void f(int? i, int? j) {
  if (i == null) return; // (1)
  // ... Additional code ...
  print(i.isEven); // (2) OK
  var foo = () {
    i = j;
  };
  // ... Use foo ...
}

Another option is to create a local variable, so it isn't write captured:

void f(int? i, int? j) {
  var foo = () {
    i = j;
  };
  // ... Use foo ...
  var i2 = i;
  if (i2 == null) return; // (1)
  // ... Additional code ...
  print(i2.isEven); // (2) OK because `i2` isn't write captured.
}

Or you can do a redundant check:

void f(int? i, int? j) {
  var foo = () {
    i = j;
  };
  // ... Use foo ...
  if (i == null) return; // (1)
  // ... Additional code ...
  print(i!.isEven); // (2) OK due to `!` check.
}

The cause: The variable is written to outside of a closure or function expression, and the type promotion location is inside the closure or function expression.

Example:

void f(int? i, int? j) {
  if (i == null) return;
  var foo = () {
    print(i.isEven); // (1) ERROR
  };
  i = j;             // (2)
}

Flow analysis reasons that there's no way to determine when foo might get called, so it might get called after the assignment at (2), and thus the promotion might no longer be valid. As with loops, this demotion happens regardless of the type of the right hand side of the assignment.

Solution:

A solution is to create a local variable:

void f(int? i, int? j) {
  if (i == null) return;
  var i2 = i;
  var foo = () {
    print(i2.isEven); // (1) OK because `i2` isn't changed later.
  };
  i = j; // (2)
}

Example:

A particularly nasty case looks like this:

void f(int? i) {
  i ??= 0;
  var foo = () {
    print(i.isEven); // ERROR
  };
}

In this case, a human can see that the promotion is safe because the only write to i uses a non-null value and happens before foo is ever created. But flow analysis isn't that smart.

Solution:

Again, a solution is to create a local variable:

void f(int? i) {
  var j = i ?? 0;
  var foo = () {
    print(j.isEven); // OK
  };
}

This solution works because j is inferred to have a non-nullable type (int) due to its initial value (i ?? 0). Because j has a non-nullable type, whether or not it's assigned later, j can never have a non-null value.

The cause: The variable you're trying to promote is write captured outside of a closure or function expression, but this use of the variable is inside of the closure or function expression that's trying to promote it.

Example:

void f(int? i, int? j) {
  var foo = () {
    if (i == null) return;
    print(i.isEven); // ERROR
  };
  var bar = () {
    i = j;
  };
}

Flow analysis reasons that there's no way of telling what order foo and bar might be executed in; in fact, bar might even get executed halfway through executing foo (due to foo calling something that calls bar). So it isn't safe to promote i at all inside foo.

Solution:

The best solution is probably to create a local variable:

void f(int? i, int? j) {
  var foo = () {
    var i2 = i;
    if (i2 == null) return;
    print(i2.isEven); // OK because i2 is local to this closure.
  };
  var bar = () {
    i = j;
  };
}

Unless stated otherwise, the documentation on this site reflects Dart 3.5.4. Page last updated on 2024-06-21. View source or report an issue.