access modifier

[Gunj Agarwal]

public class poly{
public static void main(String a[]){
A ref1=new c();
B ref2=(B)ref1;
System.out.println(ref2.g());
}
}

class A{
private int f(){ return 0;}
public int g(){ return 3;}
}

class B extends A{
public int f(){ return 1;}
public int g(){ return f();}
}


class c extends B{
public int f(){ return 2;}

}

this program prints 2 but if i change the access modifier of f() in class B
to private it prints 1....whereas for anyother it prints 2...how is it being interpreted by compiler

[Vinayagar Karpagam]

As far as I know,

when f() in B is coderanch, f() in C is interpreted as overriding f() in B.
so, the f() in C is called & 2 is output.

when f() in B is changed to private, f() in C is interpreted as "new method" (not overriding). so, the f() in B is called & 1 is output.

[Sreeram Desigan]

Dear Gunj Agarwal,
The statement System.out.println(ref2.g()); calls the g() method in B class because ref2 is a reference of B and g() is not overriden in C. Now inside g() method we have the statement return f(); Now the call to f() can be considered as this.f() since this here denotes a class C object the java runtime checks if f() is overriden in C. Now since f() is overriden in C the method in C is called which returns 2 and 2 is printed.
Now if you make f() in B as private then the answer is 1 because private methods cannot be overriden. Hence the call return f(); in method g() of B will call the f() method in B itself. For anyother access modifiers of f() in B such as protected or default or public f() can be overriden.

[b panigrahy]

when private access modifier is used -- overriding concept does not come into picture.This is bcoz private fields or methods are visible only inside the class where they are defined. so here in class c we are not overriding f() rather a new function is defined with same name i.e. f().

so during evaluation of the method (at run time only the f() in B class is called). is it clear?