doubt in polymorphism

[rosal saai]

in polymorphism, consider dog class extends from animal class

Animal a= new Dog();

and when a.play() is called, whether it will call animal class mehod or dog's method? it depends on reference variable type?

[grigoris philippoussis]

if you've overridden play() in Dog, it will call that. otherwise it will call the base class method.
the compiler looks at the reference type, at run-time, it looks at the object.

[Tuna Töre]

Moreover,


With animal reference to dog you can only call methods which are defined in animal class only, you can not call dog's methods with animal reference to dog. You can call dogs methods with dog reference
I mean

the play() method must be defined in animal, otherwise if it were in dog class you cannot say

animalReferencetoDog.play() <----it must be defined in animal class


[ October 24, 2008: Message edited by: Anut Walidera ]

[Jesper de Jong]

Rosal, why don't you write a small program to find out the answer?

Writing small programs and experimenting yourself is a great way to learn - much better than just learning theoretical answers by heart.

[Wouter Oet]

Remember that play must be a instance method. If it is static you will call play in Animal:

public class staticTest{
public static void main(String[] args){
Animal d = new Dog();
d.play();
}
}
 
class Animal{
public Animal(){}

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

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

D:\java>javac staticTest.java
D:\java>java staticTest
Animal

[krishna pr]

Ciao Anut,


With animal reference to dog you can only call methods which are defined in animal class only, you can not call dog's methods with animal reference to dog.

I agree with you, but what if the method is overridden in the Dog class, Can't the animal reference be used then to call the overridden method in the Dog class.

[Wouter Oet]

No that is not posible.

public class AnimalTest{
       public static void main(String[] args){
               Animal[] animals = new Animal[2];
               animals[0] = new Dog();
               animals[1] = new Cat();
 
               for(Animal a: animals){
                       a.play();
               }
 
               animals[0].bark();   // <-- Compiler error
       }
}
 
class Animal{
       public Animal(){}
 
       public void play(){
               System.out.println("Animal");
       }
}
 
class Dog extends Animal{
       public Dog(){}
 
       public void play(){
               System.out.println("Dog");
       }
 
       public void bark(){
               System.out.println("Bark");
       }
}
 
class Cat extends Animal{
       public Cat(){}
 
       public void play(){
               System.out.println("Cat");
       }
}

But why not write a program yourself to find out? It's easier, faster and you learn much more from it.

[Jimmy Clark]

With animal reference to dog you can only call methods which are defined in animal class only, you can not call dog's methods with animal reference to dog.

I agree with you, but what if the method is overridden in the Dog class, Can't the animal reference be used then to call the overridden method in the Dog class.

If the method is overriden in the Dog class, that implies that the method is declared in the Animal class. Or else you would not be overriding the method. Yes, the Animal reference can be used to call the overridden method in the Dog class. The Animal reference cannot be used to call a method that is only declared and implemented in the Dog class.

With an Animal reference, you can only call methods that are declared in the Animal class. To understand this better, if the Dog class has any methods that are not declared in the Animal class, you will not be able to call these methods with an Animal reference, even if the actual object is an instance of Dog.
[ October 28, 2008: Message edited by: James Clark ]