Java中的超类和子类 [英] Superclass and Subclass in Java

问题描述

对你来说，这可能很简单，但我不知道有什么区别.我只是想知道这两个代码之间的区别.假设我有一些代码，如下所述.

For you, this might be very simple but I have no idea what the difference is.I just want to know the difference between these two codes. Suppose I have some codes as described below.

第一个类是Animal，它将成为超类

The first class is Animal which will be the Superclass

public class Animal {

    private String name;
    private int weight;
    private String sound;

    public void setName(String name){
        this.name = name;
    }

    public String getName(){
        return name;
    }

    public void setWeight(int weight){
        if(weight > 0){
            this.weight = weight;
        } else {
            System.out.println("Weight must be bigger than 0");
        }
    }

    public int getWeight(){
        return weight;
    }

    public void setSound(String sound){
        this.sound = sound;
    }

    public String getSound(){
        return sound;
    }
}

第二个类是 Dog ，它扩展了类 Animal

The second class is Dog which extends the class Animal

public class Dog extends Animal {

    public void digHole(){
        System.out.println("Dig a hole");
    }

    public Dog(){
        super();

        setSound("bark");
    }
}

最后一个类是WorkWithAnimals，它将打印输出

The last class is the WorkWithAnimals which will print the output

public class WorkWithAnimals {

    public static void main(String args[]){

        Dog fido = new Dog();

        fido.setName("Dadu");
        System.out.println(fido.getName());

        fido.digHole();
        fido.setWeight(-1);
    }
}

我的问题是，Animal fido = new Dog() 和 Dog fido = new Dog() 有什么区别?

My question is, what is the difference between Animal fido = new Dog() and Dog fido = new Dog() ?

既然Dog已经继承了Animal，为什么还要写Animal fido = new Dog()这样的代码?

Since Dog already extends Animal, why do we have to write the code like Animal fido = new Dog() ?

它们都打印出相同的结果，不是吗?

Both of them print the same result, don't they?

推荐答案

Animal"类将不包含 digHole，所以你应该得到一个编译时错误；因此:

The class "Animal" will not include digHole, so you should get a compile time error; thus:

Animal fido = new Dog();
fido.dighole(); //<--compile time error

然而，

Dog fido = new Dog();
fido.dighole(); 

会好的.

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