Overidden functions

[meeta verma]

I couldn't understand properly the ans to this quest.

class Parent
{
private void method1()
{
System.out.println("Parent's method1()");
}


public void method2()
{
System.out.println("Parent's method2()");
method1();
}

}


class Child extends Parent
{
public void method1()
{
System.out.println("Child's method1()");
}


public static void main(String args[])
{
Parent p = new Child();
p.method2();
}
}

Ans o/p is
Parent's method2();
Parent's method1();
I understand that private methods are not inherited and not overridden. Still i was expecting "Child's method1()" to be printed. When p.method2()(then inheritted method2)is called, a this pointer is passed which points to Child's object. Child doesn't know anything about the super classes method1(), but then the ans......I think I am making a blunder somewhere. Pls help me in this.
Also if u could give some link on this...

[Jacky Chow]

hi meeta,
Because the Child does not override the method1() in Parent, method1() in Child class and method1() in Parent is two different methods, so, the Parent.method2() just call it's(class Parent's) own method1(), this is because the method1() exists in Parent and it is not be overrided;

[Deepa Korecherla]

Since the private method cannot be overridden, the two methods act as two individual methods...Try putting protected in place of private ..then the result u expected will come.
Try reading about accessibility and overridden methods topics.