not_assigned_potentially_non_nullable_local_variable

The non-nullable local variable '{0}' must be assigned before it can be used.

The analyzer produces this diagnostic when a local variable is referenced and has all these characteristics:

• Has a type that's potentially non-nullablePotentially non-nullableA type that is either non-nullable explicitly or due to being a type parameter. Learn more.
• 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 assignmentDefinite assignmentThe determination of whether a variable has definitely been assigned a value before it's used. Learn more.

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;
}

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;
}