Dart / Flutter - class 不必实现抽象 class 的抽象方法？

Question

在 Flutter 中我们有 class

abstract class PreferredSizeWidget implements Widget {
  Size get preferredSize;
}

和

@immutable
abstract class Widget extends DiagnosticableTree {
  const Widget({ this.key });
  final Key? key;
  @protected
  @factory
  Element createElement();
  @override
  String toStringShort() {
    final String type = objectRuntimeType(this, 'Widget');
    return key == null ? type : '$type-$key';
  }

  @override
  void debugFillProperties(DiagnosticPropertiesBuilder properties) {
    super.debugFillProperties(properties);
    properties.defaultDiagnosticsTreeStyle = DiagnosticsTreeStyle.dense;
  }

  @override
  @nonVirtual
  bool operator ==(Object other) => super == other;

  @override
  @nonVirtual
  int get hashCode => super.hashCode;
  static bool canUpdate(Widget oldWidget, Widget newWidget) {
    return oldWidget.runtimeType == newWidget.runtimeType
        && oldWidget.key == newWidget.key;
  }
  static int _debugConcreteSubtype(Widget widget) {
    return widget is StatefulWidget ? 1 :
           widget is StatelessWidget ? 2 :
           0;
    }
}

我的问题是：为什么如果我们混合使用 PreferredSizeWidget class 我们不必实现所有抽象 class 小部件方法和抽象方法？

我试着模拟了一些代码

class A implements B {}

abstract class B extends C {}

abstract class C {
  void test() {
    print("test");
  }
}

出现错误

Error: The non-abstract class 'A' is missing implementations for these members:
 - C.test
class A implements B {
      ^
lib/main.dart:16:8:

Answer 1

Haven’t run it myself, but in your case A isn’t abstract anymore so it has to implement everything, in the case of PreferredSizeWidget - you usually extend/implement/ mixin it on a widget (which already implements everything from widget) thus leaving you with only preferredSize

@Norbert515的回答。

要点是，即使我们 mixin/extend PreferredSizeWidget 我们实际上实现了所有的 Widget 方法，但是因为我们通常将它混合在 Widget 上，所以我们不必隐式覆盖这些方法.

Dart / Flutter - class 不必实现抽象 class 的抽象方法？

Dart / Flutter - class doesn't have to implement abstract class's abstract methods?

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