如果我已完成向上转换，为什么要从子类重写方法调用? [英] why overridden method calling from Subclass if i have done up-casting?

问题描述

我刚刚开始学习java :: Inheritance，并且在混合使用Up-Casting时感到困惑.

i have just started learning java::Inheritance and confused while mixing Up-Casting.

class Example{
     public void methodOne(){
        System.out.println("Example::Method_1");
     }

     public void methodTwo(){
        System.out.println("Example::Method_2");
     }
}

public class Test extends Example{

     public void methodTwo(){                  //Method overriding
        System.out.println("Test::Method_2");
     }

     public void methodThree(){
        System.out.println("Test::Method_3");
     }

    public static void main(String[] args){

        Example exa = new Test();             // UpCasting

        exa.methodOne();                      // Printing Example::Method_1
        exa.methodTwo();                      // Printing Test::Method_2
        // exa.methodThree();                 // Error : can not find symbol
 }
}

有人可以解释一下，这是怎么回事?

may someone please explain, what happening here??

推荐答案

使用继承时，对您在其上调用方法的对象的引用的编译时类型仅用于(在编译时)查看方法可以被调用.

When using inheritance, the compile-time type of the reference to an object on which you call a method is only used to see (at compile time) if the method may be invoked.

但是在调用时，编译时类型是什么都没有关系.在这种情况下，真正重要的是对象的运行时类型.它是Test，因此首先在Test上搜索该方法.

But at the invocation time, it does not matter what that compile-time type is. What really matters in this case is the runtime type of the object. It is Test, so the method is searched on Test first.

对于methodOne()，它有点不同:它不会被Test覆盖，因此将调用其超类(Example)中的版本.

For methodOne() it is a bit different: it is not overriden by Test, so the version from its superclass (Example) is invoked.

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