generics - 为什么 Dart 泛型类不能根据构造函数确定其类型

Question

我正在为我的 FutureBuilder 编写自己的 MyFutureBuilder，每次都做同样的事情(除了使用我指定单独函数的快照数据)。

typedef MyFutureBuilderBuildFunction<T> = Widget Function(T snapshotData);

class MyFutureBuilder<T> extends StatefulWidget {
  final MyFutureBuilderBuildFunction<T> builder;
  final Future<T> future;

  MyFutureBuilder({this.future, @required this.builder})
      : assert(builder != null);

  @override
  State<StatefulWidget> createState() {
    return MyFutureBuilderState(future: future, builder: builder);
  }
}

class MyFutureBuilderState<T> extends State<MyFutureBuilder<T>> {
  MyFutureBuilderBuildFunction<T> builder;
  Future<T> future;

  MyFutureBuilderState({this.future, @required this.builder})
      : assert(builder != null);

  @override
  Widget build(BuildContext context) {
    return FutureBuilder<T>(
      future: future,
      builder: (context, snapshot) {
        if (snapshot.hasData) {
          return builder(snapshot.data);
        } else if (snapshot.hasError) {
          return Text('${snapshot.error}');
        }

        return Center(child: CircularProgressIndicator());
      },
    );
  }
}

当我用

实例化它时

Future<Article> _article;
MyFutureBuilder(
    future: _article,
    builder: (snapshotData) {

snapshotData 的类型是dynamic 而不是Article。不能根据 _article 的类型自动确定类型 T，Future？

当我将类型单独提供给 MyFutureBuilder 时(恕我直言，这是多余的)

Future<Article> _article;
MyFutureBuilder<Article>(
    future: _article,
    builder: (snapshotData) {

snapshotData的类型当然是Article。

Answer 1

因为 dart 是类型安全的语言。

在第一个代码中，您应该在使用前将返回的快照数据转换为您需要的类型

Future<Article> _article;
MyFutureBuilder(
  future: _article,
  builder: (snapshotData) {
    Article article = snapshotData as Article;
  }
)

或者你可以给 future builder 一个类型参数来自动转换它，就像你在第二个代码中所做的那样