如果存在具有相同名称的方法的Mixin，那么如何调用Super Class方法 [英] How to call Super Class method if there is a Mixin with a method with the same name

问题描述

代码，此处为要点，将显示 e 。如果我删除覆盖，即从 Baz 中删除​​输出，它将打印 w Bar 的code>。

这使我得出结论，方法优先级是自己的类 -> mixin -> 超类。

The code, here as a Gist, will print e. If I remove the override, i.e. remove output from Baz, it will print w from Bar.
This leads me to the conclusion that the method "priority" is own class -> mixin -> super class.

如果我添加更多的mixin，例如像这样：

If I add more mixins, e.g. like this:

mixin Zoo {
  output() {
    print('j');
  }
}

class Baz extends Foo with Bar, Zoo {
// ...

现在，输出为 j 。如果我在 Bar 和 Zoo 之间交换：

Now, the output is j. If I swap around Bar and Zoo:

class Baz extends Foo with Zoo, Bar {
// ...

现在，输出点再次是 w 。

因此，我将这样定义优先级： 自己的班级 -> 最后一次混音 -> 倒数第二个混音 -> 超一流。

Now, the outpt is w again.
Consequently, I would define the priority like this: own class -> last mixin -> nth-last mixin -> super class.

我有什么办法 control 这种行为，即即使 mixin 具有相同名称的方法，也调用超级调用方法？

Is there any way for me to control this behavior, i.e. call the super call method even when a mixin has a method with the same name?

您可能还想知道为什么我要这样做，而不仅仅是重命名方法。

振兴所有 State 的方法有 dispose 的方法，如果我有 mixin dispose 方法的c $ c>，它将破坏 State 的功能离子性，因为 mixin 的 dispose 方法具有上述优先级。

You might be askin why I would want to do this and not just rename the methods.
Well, in Flutter all State's have a dispose method and if I have a mixin that has dispose method as well, it will break the State's functionality because the mixin's dispose method takes priority as illustrated above.

super.output 也会调用mixin方法，这就是为什么不起作用。您可以尝试将以下构造函数添加到 Baz ：

super.output will call the mixin method as well, which is why that does not work. You can try adding the following constructor to Baz:

Baz() {
  super.output();
}

即使这样做有效，由于 dispose 方法是从外部调用的。

Even if this worked, it would not help as the dispose method in the Flutter case is called from the outside.

推荐答案

在mixins中-声明混合的顺序非常重要。

在将混合应用于类时，

Dart中的Mixins通过创建一个新类来工作，该类将mixin的实现层放在一个超类之上，以创建一个新类-它不是超类的侧面，而是上层，因此在解析查询中没有歧义
源

Mixins in Dart work by creating a new class that layers the implementation of the mixin on top of a superclass to create a new class — it is not "on the side" but "on top" of the superclass, so there is no ambiguity in how to resolve lookups source

class A {
  String getMessage() => 'A';
}

class B {
  String getMessage() => 'B';
}

class P {
  String getMessage() => 'P';
}

class AB extends P with A, B {}

class BA extends P with B, A {}

void main() {
  String result = '';

  AB ab = AB();
  result += ab.getMessage();

  BA ba = BA();
  result += ba.getMessage();

  print(result);
}

AB类和AB类都使用A和B混合扩展了P类，但是以不同的顺序。 A，B和P这三个类都有一个名为getMessage的方法。

Both, AB and BA classes extend the P class with A and B mixins but in a different order. All three A, B and P classes have a method called getMessage.

首先，我们调用AB类的getMessage方法，然后调用BA类的getMessage方法。

First, we call the getMessage method of the AB class, then the getMessage method of the BA class.

输出为 BA 。

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