如果我已经完成了向上转换,为什么要从子类调用重写方法?
我刚开始学习java::Inheritance,在混合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
}
}
谁能解释一下,这里发生了什么?
may someone please explain, what happening here??
推荐答案
使用继承时,对调用方法的对象的引用的编译时类型仅用于查看(在编译时)是否方法可能会被调用.
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,编译时类型是什么并不重要.在这种情况下,真正重要的是对象的运行时类型.是Test,所以先在Test上搜索方法
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() 有点不同:它不会被 Test 覆盖,因此它的超类的版本(Example)被调用.
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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