Class casting in java

Question

In my code two classes are there ,subclass extends superclass.In sub class i override superclass method. On the time of object creation i created superclass reference to sub class object its working fine. But again i converted super class reference to complete super class object but it calls sub class methodn not super class method. My assumption the output is wrong.Here my code

public class OverRiding {
    public static void main(String[] args) {
        // TODO Auto-generated method stub

        //Super class reference to sub class Object
        Dog dog = new Animal();
        dog.eat();
        dog.bow();

        //Completely converted into pure Dog class Object
        Dog d = (Dog) dog;
        d.eat();
        d.bow();
    }
}

Dog class

class Dog {
    public void eat() {
        System.out.println("Dog eat Biscuits");
    }

    public void bow() {
        System.out.println("Dog bow");
    }
}

Class Animal

class Animal extends Dog {
    public void eat() {
        System.out.println("Animal eat Food");
    }
}

And my output is

Animal eat Food
Dog bow
Animal eat Food
Dog bow

And Animal shouldn't extend Dog, because every Animal is not a Dog. It should be the other way: Dog should extend Animal, because a Dog is an Animal. — JB Nizet
– JB Nizet, Commented Oct 10, 2013 at 8:50
A cast never changes the type of an object. It only allows getting a reference of a more specific type to an object. An animal will always be an animal, whatever you try to cast it to. — JB Nizet
– JB Nizet, Commented Oct 10, 2013 at 8:51

TheLostMind · Accepted Answer · 2013-10-10 09:22:57Z

5

Your concept is not correct bro...

 class Animal extends Dog

How can Animal extend a dog? . Its like saying - "EVERY ANIMAL IS A DOG" . Which is wrong... So, it should be the other way around...

This example will only confuse you further..

Donno why nobody brought this up...

answered Oct 10, 2013 at 9:22

TheLostMind

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Comments

Maroun · Accepted Answer · 2013-10-10 08:51:01Z

3

 Dog d = (Dog) dog;

Here d is not pure Dog object. It's still of Animal type. Pure Dog object would be created via Dog d = new Dog().

edited Oct 10, 2013 at 8:51

Maroun

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answered Oct 10, 2013 at 8:50

RandomQuestion

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Suresh Atta · Accepted Answer · 2013-10-10 08:52:40Z

2

   Dog dog = new Animal();
    //Completely converted into pure Dog class Object
    Dog d = (Dog) dog;

Both Dog and d are pointing to the same reference, Since you are not creating a new object. Cast won't have an effect here.

Casting and Creating is completely different.

answered Oct 10, 2013 at 8:52

Suresh Atta

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Dawood ibn Kareem · Accepted Answer · 2013-10-10 08:51:44Z

1

In this example, you've only created one object, and it's an Animal, (which sadly, is a special type of Dog). When you call eat on this object, you're always going to get the version defined in the Animal class, not the version defined in the Dog class.

That's polymorphism. The version of the method that gets called depends on the class of the object, not the type of the variable that refers to it.

answered Oct 10, 2013 at 8:51

Dawood ibn Kareem

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Maroun · Accepted Answer · 2013-10-10 08:54:06Z

1

You assumption is false:

Completely converted into pure Dog class Object

When you are casting your Animal to a Dog, your variable d is still an Animal. It is also a Dog, but the most specific type is Animal.

You are not completely converted your variable to another instance. Your instance is still the same.

For a proof, you can try this code :

Dog d = (Dog) dog;
d.getClass(); // output Animal

edited Oct 10, 2013 at 8:54

Maroun

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answered Oct 10, 2013 at 8:53

Cyril Gandon

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