扑：如何发现对继承的小部件的依赖性？

Question

我目前正在阅读提供程序包的示例代码：

// ignore_for_file: public_member_api_docs
import 'package:flutter/foundation.dart';
import 'package:flutter/material.dart';
import 'package:provider/provider.dart';

void main() => runApp(MyApp());

class Counter with ChangeNotifier {
  int _count = 0;
  int get count => _count;

  void increment() {
    _count++;
    notifyListeners();
  }
}

class MyApp extends StatelessWidget {
  @override
  Widget build(BuildContext context) {
    return MultiProvider(
      providers: [
        ChangeNotifierProvider(builder: (_) => Counter()),
      ],
      child: Consumer<Counter>(
        builder: (context, counter, _) {
          return MaterialApp(
            supportedLocales: const [Locale('en')],
            localizationsDelegates: [
              DefaultMaterialLocalizations.delegate,
              DefaultWidgetsLocalizations.delegate,
              _ExampleLocalizationsDelegate(counter.count),
            ],
            home: const MyHomePage(),
          );
        },
      ),
    );
  }
}

class ExampleLocalizations {
  static ExampleLocalizations of(BuildContext context) =>
      Localizations.of<ExampleLocalizations>(context, ExampleLocalizations);

  const ExampleLocalizations(this._count);

  final int _count;

  String get title => 'Tapped $_count times';
}

class _ExampleLocalizationsDelegate
    extends LocalizationsDelegate<ExampleLocalizations> {
  const _ExampleLocalizationsDelegate(this.count);

  final int count;

  @override
  bool isSupported(Locale locale) => locale.languageCode == 'en';

  @override
  Future<ExampleLocalizations> load(Locale locale) =>
      SynchronousFuture(ExampleLocalizations(count));

  @override
  bool shouldReload(_ExampleLocalizationsDelegate old) => old.count != count;
}

class MyHomePage extends StatelessWidget {
  const MyHomePage({Key key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    return Scaffold(
      appBar: AppBar(title: const Title()),
      body: const Center(child: CounterLabel()),
      floatingActionButton: const IncrementCounterButton(),
    );
  }
}

class IncrementCounterButton extends StatelessWidget {
  const IncrementCounterButton({Key key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    return FloatingActionButton(
      onPressed: Provider.of<Counter>(context).increment,
      tooltip: 'Increment',
      child: const Icon(Icons.add),
    );
  }
}

class CounterLabel extends StatelessWidget {
  const CounterLabel({Key key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    final counter = Provider.of<Counter>(context);
    return Column(
      mainAxisSize: MainAxisSize.min,
      mainAxisAlignment: MainAxisAlignment.center,
      children: <Widget>[
        const Text(
          'You have pushed the button this many times:',
        ),
        Text(
          '${counter.count}',
          style: Theme.of(context).textTheme.display1,
        ),
      ],
    );
  }
}

class Title extends StatelessWidget {
  const Title({Key key}) : super(key: key);

  @override
  Widget build(BuildContext context) {
    return Text(ExampleLocalizations.of(context).title);
  }
}

当用户在IncrementCounterButton按下FloatingRadioButton ，将在CounterLabel和IncrementCounterButton上调用build() 。

它们都依赖于继承的窗口小部件，该窗口小部件已更新。 颤振如何发现这种依赖性？

我假设BuildContext是通过对Provider.of<>()的调用来修改的。 这就是为什么我们要添加IncrementCounterButton ，它本身没有功能吗？ 只是将对Provider.of<>()的调用移到其较大的父窗口小部件之外，这样重建起来会更昂贵？

Answer 1

绑定窗口小部件InheritedWidget及其使用者通过BuildContext创建。

考虑以下InheritedWidget：

class Foo extends InheritedWidget {}

然后， Foo的后代可以通过调用以下命令来订阅它：

BuildContext context
context.inheritFromWidgetOfExactType(Foo);

值得注意的是，小部件可以通过执行以下操作而无需订阅即可获取InheritedWidget：

BuildContext context
context.ancestorInheritedElementForWidgetOfExactType(Foo);

---

此调用通常由.of(context)模式在内部执行。

对于provider ，该订阅是通过调用Provider.of<T>(context) 。

provider还公开了一个可选参数，以故意不订阅继承的小部件：

T value = Provider.of<T>(context, listen: false);