flutter - 空检查不会在Dart中引起类型提升

Question

我正在升级基于Flutter框架的个人软件包。我在Flutter Text小部件源代码中注意到here，存在一个空检查:

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

但是， textSpan!仍在使用 !运算符。无需将 textSpan提升为非可空类型，而不必使用 !运算符吗？但是，尝试删除该运算符将产生以下错误:

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.

这是一个独立的示例:

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

我收到一个编译时错误:

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

或者，如果我尝试获取 _mySting.length，则会出现以下错误:

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

我以为执行空检查将把 _myString提升为非空类型。为什么不呢？
My question已在GitHub上解决，因此我在下面发布了答案。

Answer 1

Dart 工程师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.

因此，本地类型推广工作:

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

但是实例变量不会升级。为此，您需要使用 !运算符手动告诉Dart您确保实例变量在这种情况下不为null:

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