返回子类型对象而不是接口所要求的超类型对象 [英] Returning a subclass type object instead of superclass type object as demanded by interface

问题描述

我有一个名为Node的类和另一个名为ClassicNode的类，它扩展了Node。现在我有一个由ClassicNode类实现的接口AgentInterface。接口声明必须有方法

I have a class called Node and another class called ClassicNode which extends Node. Now I have an interface AgentInterface implemented by ClassicNode class. The interface states that there must be a method

Node selection();

如您所见，返回类型应为Node类型。但是在类ClassicNode中我可以像这样实现它： -

As you can see, the return type should be of type Node. But in the class ClassicNode can I implement it like this instead:-

ClassicNode selection(){
    //Code
}

这会满足界面吗？ （因为ClassicNode继承了Node）

Will this satisfy the interface? (since ClassicNode inherits Node)

推荐答案

是的;它被称为协变返回。但是请注意，你不能用参数做同样的事情;它们必须完全匹配。

Yes; it's called covariant return. Note, though, that you cannot do the same thing with parameters; they must match exactly.

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