"The operator can’t be unconditionally invoked because the receiver can be null" error after migrating to Dart null-safety

Question

I'm upgrading a personal package that is based on the Flutter framework. I noticed here in the Flutter Text widget source code that there is a null check:

if (textSpan != null) {
  properties.add(textSpan!.toDiagnosticsNode(name: 'textSpan', style: DiagnosticsTreeStyle.transition));
}

However, textSpan! is still using the ! operator. Shouldn't textSpan be promoted to a non-nullable type without having to use the ! operator? However, trying to remove the operator gives the following error:

An expression whose value can be 'null' must be null-checked before it can be dereferenced.
Try checking that the value isn't 'null' before dereferencing it.

Here is a self-contained example:

class MyClass {
  String? _myString;
  
  String get myString {
    if (_myString == null) {
      return '';
    }
    
    return _myString; //   <-- error here
  }
}

I get a compile-time error:

Error: A value of type 'String?' can't be returned from function 'myString' because it has a return type of 'String'.

Or if I try to get _mySting.length I get the following error:

The property 'length' can't be unconditionally accessed because the receiver can be 'null'.

I thought doing the null check would promote _myString to a non-nullable type. Why doesn't it?

My question was solved on GitHub so I'm posting an answer below.

Answer 1

Dart engineer Erik Ernst says on GitHub:

Type promotion is only applicable to local variables. ... Promotion of an instance variable is not sound, because it could be overridden by a getter that runs a computation and returns a different object each time it is invoked. Cf. dart-lang/language#1188 for discussions about a mechanism which is similar to type promotion but based on dynamic checks, with some links to related discussions.

So local type promotion works:

  String myMethod(String? myString) {
    if (myString == null) {
      return '';
    }
    
    return myString;
  }

But instance variables don't promote. For that you need to manually tell Dart that you are sure that the instance variable isn't null in this case by using the ! operator:

class MyClass {
  String? _myString;
  
  String myMethod() {
    if (_myString == null) {
      return '';
    }
    
    return _myString!;
  }
}

Answer 2

The Error:

Let's say, this is your code and you're doing a null check on the instance variable and still seeing an error:

class Foo {
  int? x;

  double toDouble() {
    if (x != null) return x.toDouble(); // <-- Error
    return -1;
  }
}

The method 'toDouble' can't be unconditionally invoked because the receiver can be 'null'.

The error you see in code like this is because Getters are not promoted to their non-nullable counterparts. Let's talk about the reason why.

---

Reason of the Error:

Let's say, there's a class Bar which extends Foo and overrides x field and implemented like this:

class Bar extends Foo {
  @override
  int? get x => (++_count).isOdd ? 1 : null;
  int _count = 0;
}

Now, if you do

Bar().toDouble();

You would have run into a runtime null error, which is why getters type promotion is prohibited.

---

Solutions:

We need to cast away nullability from int?. There are generally 3 ways to do this.

• Use local variable (Recommended)

  double toDouble() {
    final x = this.x; // <-- Use a local variable
    if (x != null) return x.toDouble(); 
    return -1;
  }
  

• Use ?. with ??

  double toDouble() {
    return x?.toDouble() ?? -1; // Provide a default value
  }
  

• Use null-assertion operator (!)

  You should only use this solution when you're 100% sure that the variable (x) will never be null.

  double toDouble() {
    return x!.toDouble(); // Null assertion operator
  }