14

在一个类中，与超类中的字段同名的字段会隐藏超类的字段。

public class Test {

    public static void main(String[] args) {

        Father father = new Son();
        System.out.println(father.i); //why 1?
        System.out.println(father.getI());  //2
        System.out.println(father.j);  //why 10?
        System.out.println(father.getJ()); //why 10?

        System.out.println();

        Son son = new Son();
        System.out.println(son.i);  //2 
        System.out.println(son.getI()); //2
        System.out.println(son.j); //20
        System.out.println(son.getJ()); //why 10?
    }  
}

class Son extends Father {

    int i = 2;
    int j = 20;

    @Override
    public int getI() {
        return i;
    }
}

class Father {

    int i = 1;
    int j = 10;

    public int getI() {
        return i;
    }

    public int getJ() {
        return j;
    }
}

有人可以为我解释一下结果吗？

4

2 回答 2

13

在java中，字段不是多态的。

Father father = new Son();
System.out.println(father.i); //why 1? Ans : reference is of type father, so 1 (fields are not polymorphic)
System.out.println(father.getI());  //2 : overridden method called
System.out.println(father.j);  //why 10? Ans : reference is of type father, so 2
System.out.println(father.getJ()); //why 10? there is not overridden getJ() method in Son class, so father.getJ() is called

System.out.println();

// same explaination as above for following 
Son son = new Son();
System.out.println(son.i);  //2 
System.out.println(son.getI()); //2
System.out.println(son.j); //20
System.out.println(son.getJ()); //why 10?
于 2012-09-03T09:07:12.330 回答
4

根据覆盖和隐藏方法

被调用的隐藏方法的版本取决于它是从超类调用还是从子类调用。

即，当您通过超类引用调用在子类中被覆盖的方法时，将调用超类方法并访问超类成员。

这解释了以下内容，因为使用的参考是超类：

System.out.println(father.i);  //why 1?
System.out.println(father.j);  //why 10?
System.out.println(father.getJ()); //why 10?

以下类似：

System.out.println(son.getJ()); //why 10?

因为getJ()未在版本中定义，Son所以Father调用看到Father类中定义的成员。

如果您阅读隐藏字段；他们特别不推荐这样的编码方法

一般来说，我们不建议隐藏字段，因为它会使代码难以阅读。

于 2012-09-03T09:15:24.373 回答