覆盖Java中的Vs类变量的方法重写 [英] method overriding Vs class variable overriding in java

问题描述

我只是尝试一些示例代码来检查Java中的类变量重写行为.下面是代码:

I was just trying some sample code for checking class variable overriding behavior in Java. Below is the code:

class A{
  int i=0;

  void sayHi(){
    System.out.println("Hi From A");
  }
}

 class B extends A{
  int i=2;

  void sayHi(){
    System.out.println("Hi From B");
  }
}


public class HelloWorld {
 public static void main(String[] args) {
    A a= new B();
    System.out.println("i->"+a.i); // this prints 0, which is from A
    System.out.println("i->"+((B)a).i); // this prints 2, which is from B
    a.sayHi(); //  method from B gets called since object is of type B
  }
}

我无法理解下面这两行的情况

I am not able to understand whats happening at these two lines below

System.out.println("i->"+a.i); // this prints 0, which is from A
System.out.println("i->"+((B)a).i); // this prints 2, which is from B

即使对象的类型为B，为什么a.i也打印0?为何将它投射到B之后打印2?

Why does a.i print 0 even if the object is of type B? And why does it print 2 after casting it to B?

推荐答案

i不是方法-它是数据成员.数据成员不会覆盖，它们会隐藏.因此，即使您的实例是B，它也有两个数据成员-A中的i和B中的i.当您通过A引用进行引用时，您将获得前者；而当您使用B引用(例如，通过显式转换它)时，您将获得后者.

i is not a method - it's a data member. Data members don't override, they hide. So even though your instance is a B, it has two data members - i from A and i from B. When you reference it through an A reference you will get the former and when you use a B reference (e.g., by explicitly casting it), you'll get the latter.

实例方法的行为有所不同.不管引用的类型如何，由于实例是一个B实例，因此您将获得多态行为并打印出字符串"Hi From B".

Instance methods, on the other hand, behave differently. Regardless of the the type of the reference, since the instance is a B instance, you'll get the polymorphic behavior and get the string "Hi From B" printed.

这篇关于覆盖Java中的Vs类变量的方法重写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IT屋！