Hi all,

First of all I want to thanks for Rijus & Sirje for their replies for my previors doubt. I m now cleared with that connept at some extent. But in the following program why the compiler gives error. The super class animal reference f cant invoke subclass drink method....? What is the reason...?


class animal
{
public void eat()
{
System.out.println("you are in animal");
}
}

class horse extends animal
{
public void eat()
{
System.out.println("you are in horse");
}
public void drink()
{

}
}

public class Flo
{
public static void main(String args[])
{
animal d=new animal();
animal f=new horse();
d.eat();
f.eat();
f.drink();

}
}


Thanks & Regards,
Abhijit

Hi Abhijit !

It is illegal access.
Look at this line....
animal f=new horse();

here u r crate an reference object 'f' for animal class, it holds horse class 's properties and behaviors.It called technically Polymorphic objects.

Polymorphic object access only overridden methods .
Suppose u want to access the non-overridden methods please
you can do explicit type casting.

i.e
horse h=(horse)f;then u call the method
h.drink();

Hi Abhi,

COMPILE TIME CHECK:

When we call a method using any class reference then first compiler checks whether the method present in the class? If not then it is a compile time error. As in this case

f.drink(); //Compile time Error not found in animal class

So

animal f = new horse();

Here f can be used to call only overridden methods.reason see here-

RESOLVING CALLS TO OVERRIDDEN METHODS:

Call to overridden method is resolved at run time.
Which method to call is determined on the basis of the object not on the basis of the reference.

d.eat(); //since eat() is present in super class (animal) so it compiles but calls super class method because object is of super class

f.eat(); //since eat() is present in super class (animal) so it compiles but calls the sub class method

Above we saw that in both the cases the presence of eat() method is being checked in super class at compile time because both references(f and d) are of super class.

Now which method to call is resolved at run time on the basis of the object the reference refer to

So if you want to call the drink() method using same mechanism as you are calling then define it in super class also to avoid compile time check.

Hope it would help you.