本文介绍了在Java中实例化接口的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有这个界面:

public interface Animal {
    public void Eat(String name);
}

此代码实现了界面:

public class Dog implements Animal {
    public void Eat(String food_name) {
        System.out.printf(food_name);
    }

    public static void main(String args[]) {
        Animal baby2 = new Dog(); //HERE!!!!!!!!!!!!!!!!!!!!!!
        baby2.Eat("Meat");
    }
}

我的问题是,为什么代码有效?无法实例化接口。然而在这种情况下,界面被实例化(标记为HERE !!!!!!!!!!!!!!)。

My question is, why does the code work? An interface cannot be instantiated. Yet in this case, interface was instantiated (marked with the comment "HERE!!!!!!!!!!!!!").

这里发生了什么?

推荐答案

不,不是 - 你正在实例化一个 Dog ,但由于 Dog Animal ,您可以将变量声明为 Animal 。如果您尝试实例化接口 Animal ,它将是:

No it is not - you are instantiating a Dog, but since a Dog is an Animal, you can declare the variable to be an Animal. If you try to instantiate the interface Animal it would be:

Animal baby2 = new Animal();

试试看,并惊恐地看着编译器尖叫:)

Try that, and watch the compiler scream in horror:)

这篇关于在Java中实例化接口的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

09-12 08:06