空检查不会在Dart中引起类型提升

问题描述

我正在升级基于Flutter框架的个人软件包.我注意到这里在Flutter Text小部件源代码中，存在空检查:

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));
}

但是， textSpan！仍在使用！运算符.是否不必须使用！运算符将 textSpan 提升为不可为空的类型?但是，尝试删除该运算符会出现以下错误:

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.

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:

I get a compile-time error:

或者，如果我尝试获取 _mySting.length ，则会出现以下错误:

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

我认为执行null检查将把 _myString 提升为不可为null的类型.为什么不呢?

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

我的问题已在GitHub上解决，因此我发布了答案在下面.

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

推荐答案

飞镖工程师Erik Ernst 在GitHub上的说法:

Dart engineer Erik Ernst says on GitHub:

因此本地类型推广有效:

So local type promotion works:

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

    return myString;
  }

但是实例变量不提升.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:

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!;
  }
}

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