which version of method is actually called ? the reference one or actual object one ?

[A Agr]

In K&B book,there is an example

public class TestAnimals {
	public static void main(String[] args) {
		Animal a = new Animal();
		Animal b = new Horse(); // Animal ref, but a Horse object
		a.eat(); // Runs the Animal version of eat()
		b.eat(); // Runs the Horse version of eat()
	}
}

class Animal {
	public void eat() {
		System.out.println("Generic Animal Eating Generically");
	}
}

class Horse extends Animal {
	public void eat() {
		System.out.println("Horse eating hay, oats, " + "and horse treats");
	}

	public void buck() {
	}
}

And then it says,

Animal c = new Horse();
c.buck(); // Can't invoke buck();
// Animal class doesn't have that method

Why will it not check for Horse class for buck method ?

Can somebody help me out in making clear picture when methods on reference variable are called and when methods on actual object is called ?

[akash azal]

in case of overriding method on object type is called and in case of overloading method on reference type is called

[Balu Sadhasivam]

Animal c = new Horse();
c.buck(); // Can't invoke buck();
// Animal class doesn't have that method

Why will it not check for Horse class for buck method ?

I remember sometime ago seen similar post and the reply impressed me was

" Animal reference cannot used to call buck() method " because simply there is no button( buck) in the remote (Animal) to use

[Rajdeeep Singh]

A Agr wrote:

Animal c = new Horse();
c.buck(); // Can't invoke buck();
// Animal class doesn't have that method

Why will it not check for Horse class for buck method ?

Can somebody help me out in making clear picture when methods on reference variable are called and when methods on actual object is called ?

Compiler will try to look for a buck() method in Animal and will not find it. So, this will not even compile

[Mumtaz Khan]

The Reference variable's type not the object's type determines which method can be called and polymorphic method invocation apply only to overridden method. Thats why this code

 Animal b = new Horse(); // Animal ref, but a Horse object 
b.eat(); // Runs the Horse version of eat() 
}    

runs eat() method of Horse.
As eat() method is an overridden method and Polymorphism applies to overriding so object's type( here Horse) method runs.

 Animal c = new Horse(); 
c.buck(); // Can't invoke buck();    

Here the reference type is Animal which doesn't have buck() method so it will not compile. buck() is not an overridden method but an instance method of Horse. Calling of method is determined at compile time based on reference type

[Stephen Davies]

Animal reference cannot used to call buck() method " because simply there is no button( buck) in the remote (Animal) to use

This is a great explanation, it looks like a Sierra & Bates Head First quote!

[Jesper de Jong]

A Agr, welcome to JavaRanch.

Please use code tags when you post source code.

[A Agr]

Alright. I didn't know before , will use code tags from now.