"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

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? i = 0;

  double func() {
    if (i != null) return i.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 override i variable and doesn't assign it any value (keeping it null):

class Bar extends Foo {
  @override
  int? i;
}

So, if you could do

print(Bar().func() * 2);

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 (more ways include the use of as, is, etc)

• Use local variable (Recommended)

  double bar() {
    var i = this.i; // <-- Use of local variable.
    if (i != null) return i.toDouble();
    return -1;
  }
  

• Use ?. with ??

  double bar() {
    return i?.toDouble() ?? -1; // Provide some default value.
  }
  

• Use Bang operator (!)

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

  double bar() {
    return i!.toDouble(); // <-- Bang operator in play.
  }
  

Answer 2

style: Theme.of(context).textTheme.headline5!.copyWith(

style: Theme.of(context).textTheme.headline5!.copyWith(
                        color: Colors.white

Try making the call conditional using ? or a null safety checker - !