Calling a Overriding method in a abstract class

[Sivaraman Lakshmanan]

Hi all,
I have basic question which confused me. I am pasting the code below

    
public abstract class A {
 
    public void doit(){
        System.out.println("This method is in A..");
    }
 
    public abstract void tryit();
 
}
 
public class B extends A{
 
    public static void main(String[] args) {
 
      B b = new B();
      b.doit(23);
 
      A a = new B();
      a.doit(23);
    }
 
    public void doit(){
       System.out.println("This is a method in A which is Overriden");
    }
 
    public void tryit() {
        System.out.println("tryit in B implemented");
    }
 
}
 

Here the doit() method is overridden by class B. Now how do I access the doit() method of class A, from class B.

Thanks in Advance,

Regards,
Sivaraman.L

[Sivaraman Lakshmanan]

Hi All,
In the previous post the class B code is wrong so I am posting it again.

 
public class B extends A{
 
    public static void main(String[] args) {
 
      B b = new B();
      b.doit();
 
      A a = new B();
      a.doit();
    }
 
    public void doit(){
       System.out.println("This is a method in A which is Overriden");
    }
 
    public void tryit() {
        System.out.println("tryit in B implemented");
    }
 
}

Regards,
Sivaraman.L

[Jeff Albertson]

It's unclear what you intend this to mean: "Now how do I access the doit() method of class A, from class B."

1. Are you trying to invoke A's implementation of doit as a subroutine from within one of B's methods, say its version of doit? Use super:

//Try this in B:
public void doit(){
    super.doit();
    System.out.println("This is a method in A which is Overriden");
}

2. Or are you confused why the following give the same results?

B b = new B();
b.doit();
//versus...
A a = new B();
a.doit();

In both cases, B's implementation of doit is invoked because the object being passed the message "doit" is an instance of B (because you wrote new B() twice).

[Ankur Sharma]

Actually What happen every time when you write such a code
e.g.

Class Base {
 
public void baseMethod(){
}
 
}
 
Class Derived extends Base{
public void derivedMethod(){
}
public void baseMethod(){
}
Derived d = new Base();
d.baseMethod();
 
}

It will search automatically in Base Class first and if there present then execute that method otherwise it checks the method declaration in Derived Class. This is the way of Execution. So this is a known behaviour of JAVA. SO don't be surpriced too much.
.

Hope You will be understand this.

[Garrett Rowe]

It will search automatically in Base Class first and if there present then execute that method otherwise it checks the method declaration in Derived Class...

Thats actually the opposite of what will happen.

[Sivaraman Lakshmanan]

Hi all,
Thanks for your replies. I want to call the A's doit() implementation from class B but not from any method in B, but from the B's main method.
something like...

   
public static void main(String[] args) 
{
   B b = new B(); 
   b.super.doit(); // this gives a compilation error..
}

Regards,
Sivaraman.L

[Ajay A Patil]

You can do it as follows:

public static void main(String[] args) 
{
   B b = new B();    
   ((A) b).doit(); // Cast B to A
}

OR

public static void main(String[] args) 
{
   B b = new B();    
   A a = b;
   a.doit(); 
}

[Layne Lund]

Originally posted by Ajay A Patil:
You can do it as follows:

public static void main(String[] args) 
{
   B b = new B();    
   ((A) b).doit(); // Cast B to A
}

OR

public static void main(String[] args) 
{
   B b = new B();    
   A a = b;
   a.doit(); 
}

This will not produce the results requested. Both examples will still call B's doit() method. In fact, I do not think there is any way to call A's doit() method from main unless you have an instance of A. An instance of B will always invoke B's doit() method, even if you have a reference to an A pointing to it.

Layne

[Shrinivas Mujumdar]

Yoy are right Layne,
You can call A's doit() either as a super.doit() in a method or with instance of A but in this case as A is abstract 2nd option is gone.

Shrinivas