overriding : why "Superclass" is printed?

[Ramakrishnan Ponmudi]

class superclass {
private void method1() {
System.out.println("Superclass");
}

public void method2() {
method1();
}
}


class subclass extends superclass {
public void method1() {
System.out.println("Subclass");
}

public static void main(String args[]) {
subclass sub = new subclass();
sub.method2();
}
}

[Mani Ram]

You are calling method2() on subclass object.
Since the subclass doesn't have any method called method2(), the method2() of super class is called. When the method2() makes a call to method1(), it is the method1() of the super class which is being called actually. That call won't invoke the method1() in the subclass.

On the otherhand, if you try to call the method1() from your main method, you can see Subclass being printed.

[Arvind Sampath]

I observed this fact while trying to answer your question.

If the access modifier for method1() in SuperClass is changed to public,default or protected "Subclass" is printed. If the access modifier is private then "Superclass" is printed.

Can someone clarify this ?

Thanks!

Regards,
Arvind

[Ramakrishnan Ponmudi]

if u remove private access specifier , then it display "Subclass"?
how is it happen?
can u clarify.

Originally posted by Mani Ram:
You are calling method2() on subclass object.
Since the subclass doesn't have any method called method2(), the method2() of super class is called. When the method2() makes a call to method1(), it is the method1() of the super class which is being called actually. That call won't invoke the method1() in the subclass.

On the otherhand, if you try to call the method1() from your main method, you can see Subclass being printed.

[Ilja Preuss]

Simple: private methods aren't known in subclasses, and therefore consequently can't be overridden - they are just shadowed. With other words, private methods aren't polymorphic.

[Arvind Sampath]

Thanks a Lot, Ilja !

You've made things clear.


Regards,
Arvind