Null Problem

[Ramesh Sahu]

code:
--------------------------------------------------------------------------------


class AQ{
public void math(Object o){
System.out.println("ob ver");
}
public void math(String o){
System.out.println("str ver");
}
public static void main(String str[]){
AQ q=new AQ();
q.math(null);
}
}


--------------------------------------------------------------------------------

when I run the above code the output is
str ver

why?
[ July 31, 2008: Message edited by: Ramesh Sahu ]

[Seetharaman Venkatasamy]

-----------------------------------------------------------------
Java choose nearly(closest) matching method for the reference
-----------------------------------------------------------------

[Ramesh Sahu]

how can a null be matched to anything else?

[Harshit Rastogi]

when I run the above code the output is
str ver

Hi
I guess you are aware with the widening concept. In which if there are methods
void m (int i)
void m (Integer i)
void m (Object o)
void m(Object... args)

if an int i =5 is passed

so the preference wiil be given in the order
primitive type
wrapper class
object
varArgs

So considering in your case, method with String parameter will be called

[Harshit Rastogi]

how can a null be matched to anything else?

you can define

String str = null;

I guess this will answer the question.

[Sandeep Bhandari]

Just to add, its working the same way with 1.4 and 1.3 too.

class AQ{
public void math(Object o){
System.out.println("ob ver");
}
public void math(String o){
System.out.println("str ver");
}
public void math(Number o){
System.out.println("str ver");
}
public static void main(String str[]){
AQ q=new AQ();
q.math(null);
}
}   

makes the call to math method ambiguous

and the following code

 class AQ{
public void math(Object o){
System.out.println("ob ver");
}
public static void main(String str[]){
AQ q=new AQ();
q.math(null);
}
}
   

results in the output of

"ob ver"

Hope that makes some sense
[ July 31, 2008: Message edited by: Sandeep Bhandari ]

[Sandeep Bhandari]

also guess the output of following program

 class AQ{
public void math(Object o){
System.out.println("ob ver");
}
public void math(String o){
System.out.println("ob ver");
}
public static void main(String str[]){
AQ q=new AQ();
AQ a=null;
q.math(a);
}
}   

[Seetharaman Venkatasamy]

Originally posted by Sandeep Bhandari:
System.out.println("ob ver");

Hai, you are giving same message in both a case...anyway here object reference will take precedence