A class extending a class and implementing an interface with same methods ?

[K Srinivasan]

if a class extends another class and implements an interface both with the same method , which function will it implement ??

eg:

interface a
{
 public void show();
}

abstract class b
{
 abstract public void show();
}

public class c extends b implements a
{
public void show()
 {
  System.out.println("hi");
 }

public static void main(String[] args)
 {
   c t = new c();
   t.show();
 }
}

which show() does c implement ??

thanks.

[Sebastian Janisch]

The important thing is that the rules for the implementation of interfaces and abstract classes have been obeyed.

So, if you do

a myA = new c();

myA.show();

or

b myB = new c();

myb.show();

both will do the same thing...

[John de Michele]

'Albus':

Aren't you dead? You probably should change your username to your real name. Fake names are against the rules, and the moderators will get on your case.

John.

[David Newton]

albus dumbeldore wrote:which show() does c implement ??

From a pragmatic standpoint, what's the difference? If the signatures are the same, the implementing subclass will be an instance of both the superclass and interface (unless I'm misunderstanding something).

[Jd Sethi]

See java follows bottom to up approach but in this code you are overriding the same method show() in main. So it doesn't matter here which show() will be implemented here as it is defined in main class but according to java basics the show function of class B will be executed here as it is bottom to class A.

[Junilu Lacar]

Jd Sethi wrote:according to java basics the show function of class B will be executed here as it is bottom to class A.

Nope.

b.show() is declared as abstract, therefore it has no implementation, therefore there is no b.show() to execute. Same deal with a.show(). They both are just declarations and cannot be executed. A subtype must provide the implementation and in this case, that's c.show(), therefore the only executable show() is c.show().