How vararg works with overriding?

[Ravi Nistala]

Can anybody explain why i am getting the following?I am getting the answer as "A.m1" as opposed "b.m1",but when i remove the vaar args then i get "b.m1".In the K&B book there is a lot for overload and vararg but not much for overiding.or am i missing something.

public class B extends A {
 
public static void main(String[] args) {

A a = new B();
B b = (B) a;
a.m1("test");
b.m1("test");

}
 
public void m1(String a) {
System.out.println("In b.m1");
}
 

}
 
class A {
public void m1(String... a) {
System.out.println("In A.m1");
}
 

}

[ December 24, 2006: Message edited by: Ravi Nistala ]

[Aniket Patil]

Please note that

    public void m1(String a) 

is not a valid override of

  public void m1(String... a) 

since the parameter types are not same. You are overloading method m1().

Since the method to be called is decided on the reference type, a.m1() calls the method in class A and b.m1() calls the method in class B.

but when i remove the vaar args

You mean the method signature in class A becomes

   public void m1(String a) 

This causes the method in class B to override the one in class A and by the rules of dynamic dispatch, you get the method in class B getting executed for both the calls.
[ December 25, 2006: Message edited by: Aniket Patil ]

[Ravi Nistala]

Thanks Aniket for the explanation.