Question

我有一个通用的静态方法，如下所示：

static build<K>() {
    return (GenericClass<K> param) => MyClass<K>(param);
}

到目前为止，我已经尝试过：

typedef F = MyClass<K> Function(GenericClass<K> param);

但是它说：

The return type '(GenericClass<K>) → MyClass<K>' isn't a '(GenericClass<dynamic>) → MyClass<dynamic>', as defined by the method 'build'.

和

typedef F = SimpleViewModel<K> Function<k>(Store<K> param);

说：

The return type '(GenericClass<K>) → MyClass<K>' isn't a '<K>(GenericClass<K>) → MyClass<K>', as defined by the method 'build'.

MyClass看起来像这样：

class MyClass<T> {
  final GenericClass<T> param;

  MyClass(this.param);


  static build<K>() {
      return (GenericClass<K> param) => MyClass<K>(param);
  }
}

那么，有效的typedef是什么？

Answer 1

关于typedef，有两个“泛型”概念。 typedef可以是某个类型的泛型，或者typedef可以引用泛型函数（或两者）：

在T上通用的typedef：

typedef F<T> = T Function(T);

然后使用：

F first = (dynamic arg) => arg; // F means F<dynamic>
F<String> second = (String arg) => arg; // F<String> means both T must be String

typedef，它引用M上的泛型函数：

typedef F = M Function<M>(M);

然后使用：

F first = <M>(M arg) => arg; // The anonymous function is defined as generic
// F<String> -> Illegal, F has no generic argument
// F second = (String arg) => arg -> Illegal, the anonymous function is not generic

或全部：

typedef F<T> = M Function<M>(T,M);

以及用法：

F first = <M>(dynamic arg1, M arg2) => arg2; // F means F<dynamic> so the T must be dynamic
F<String second = <M>(String arg1, M arg2) => arg2; // The T must be String, the function must still be generic on the second arg and return type

如何对通用函数进行类型定义？

1 个答案: