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 是 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:)
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