需要Java代码片段输出说明

我的代码是:

class Foo {
  public int a=3;
  public void addFive() {
    a+=5;
    System.out.print("f ");
  }
}

class Bar extends Foo {
  public int a=8;
  public void addFive() {
    this.a += 5;
    System.out.print("b ");
  }
}

public class TestClass {
  public static void main(String[]args) {
    Foo f = new Bar();
    f.addFive();
    System.out.println(f.a);
  }
}

输出:

b 3

请解释一下，为什么这个问题的输出是b 3"而不是b 13"，因为该方法已被覆盖?

Please explain to me, why is the output for this question "b 3" and not "b 13" since the method has been overridden?

推荐答案

你不能在 Java 中覆盖变量，因此你实际上有两个 a 变量 - 一个在 Foo 和Bar 中的一个.另一方面 addFive() 方法是多态的，因此它修改 Bar.a (Bar.addFive() 被调用，尽管是静态类型f 是 Foo).

You cannot override variables in Java, hence you actually have two a variables - one in Foo and one in Bar. On the other hand addFive() method is polymorphic, thus it modifies Bar.a (Bar.addFive() is called, despite static type of f being Foo).

但最终您访问 f.a 并且在编译期间使用已知类型的 f 解析此引用，即 Foo.因此 Foo.a 从未被触及.

But in the end you access f.a and this reference is resolved during compilation using known type of f, which is Foo. And therefore Foo.a was never touched.

顺便说一句，Java 中的非最终变量应该从不公开.

BTW non-final variables in Java should never be public.