Dart Flutter：功能未执行

Question

我是 Dart 和 Flutter 的新手，我正在尝试测试 Future 功能。我写了一个简短的 async 函数，其中嵌套了两个 Future 对象。 问题： 函数不执行，当我尝试将其归因于变量时出现错误 This expression has a type of 'void' so its value can't be used. 和 Only static members can be accessed in initializers.。这是代码：[![在此处输入图片描述][1]][1]

  @override
  _ChoseLocationState createState() => _ChoseLocationState();
}

class _ChoseLocationState extends State<ChoseLocation> {
  int counter = 0;
  
  void simulateRequest() async {

    // first future holds family name
    String famNameFunc = await Future.delayed(Duration(seconds: 2), (){
      String famName = 'Shanshi';
      return famName;
    });


    // second future holds first name
    String compName = await Future.delayed(Duration(seconds: 1), (){
      String fstName = 'Yoshi';
      String compName = '$fstName - $famNameFunc';
      return compName;
    });

    print(compName);
  }

  dynamic funcex = simulateRequest();

  @override
  void initState() {
    super.initState();
    print('This is the initial state.');
  }
  
  
  @override
  Widget build(BuildContext context) {
    print('This is the build function processing.');
    return Scaffold(
        appBar: AppBar(
          backgroundColor: Colors.red,
          title: Text('Set Location'),
          centerTitle: true,
          ),
        body: RaisedButton(
          onPressed: (){
          setState(() {
            counter++;
          });
        },
          color: Colors.blue,
          child: Text('Counter is: $counter'),
        ),
      );
  }
}```

[1]: https://i.stack.imgur.com/XmvY9.png

Answer 1

尝试以下操作：

class ChoseLocation extends StatefulWidget {
  ChoseLocation({Key key, this.title}) : super(key: key);

  final String title;

  @override
  _ChoseLocationState createState() => _ChoseLocationState();
}

class _ChoseLocationState extends State<ChoseLocation> {
  int counter = 0;
  
  void simulateRequest() async {

    // first future holds family name
    String famNameFunc = await Future.delayed(Duration(seconds: 2), (){
      String famName = 'Shanshi';
      return famName;
    });


    // second future holds first name
    String compName = await Future.delayed(Duration(seconds: 1), (){
      String fstName = 'Yoshi';
      String compName = '$fstName - $famNameFunc';
      return compName;
    });

    print(compName);
  }

//   dynamic funcex = simulateRequest();

  @override
  void initState() {
    super.initState();
    simulateRequest();
    print('This is the initial state.');
  }
  
  
  @override
  Widget build(BuildContext context) {
    print('This is the build function processing.');
    return Scaffold(
        appBar: AppBar(
          backgroundColor: Colors.red,
          title: Text('Set Location'),
          centerTitle: true,
          ),
        body: RaisedButton(
          onPressed: (){
          setState(() {
            counter++;
          });
        },
          color: Colors.blue,
          child: Text('Counter is: $counter'),
        ),
      );
  }
}

您需要在方法示例 initState 中调用 simulateRequest()，并且由于它没有 return 任何内容，因此您无法将其分配给变量。

Answer 2

simulateRequest 有一个 return 类型的 void，所以如果你想把这个函数存储在变量中，你不应该加上括号。如果您使用括号，您将成为运行函数并将其 returned 值分配给变量 funcex，这就是为什么您得到：This expression has a type of 'void' so its value can't be used.

      @override
  void initState() {
    dynamic funcex = simulateRequest;
    funcex();
    super.initState();
    print('This is the initial state.');
  }

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