Completely stumped at this

[Parag Kale]

Hi!
I found this on one of the mock exams.
After the following code is run, the output is 10,0,20 and that is also supposed the be the right answer. But i think the answer should be 10, 0, 30. Can somebody explain what really happens to the value of v.i, in detail ?

class ValHold {
 public int i=10; 
}
 
public class obParm {
 
public static void main(String argv[]) {
 obParm o = new obParm() ;
 o.amethod();
}
// end main
 
public void amethod() {
 int i =99 ;
 ValHold v = new ValHold() ;
 v.i=30;
 
 another(v,i);
 
 System.out.println(v.i) ;
} file://end of method
 
public void another(ValHold v, int i) {
 
 i=0 ;
 v.i=20 ;
 
 ValHold vh = new ValHold() ;
 
 v= vh ;
 
 System.out.println(v.i + " " + i) ;
 
}// End of method
}//end of class

Thanx,
Parag

[Rebecca Pickett]

Originally posted by Parag Kale:
Hi!
I found this on one of the mock exams.
After the following code is run, the output is 10,0,20 and that is also supposed the be the right answer. But i think the answer should be 10, 0, 30. Can somebody explain what really happens to the value of v.i, in detail ?

Parag,
You have to walk through it carefully you'll get tripped up. I think here's what's confusing you: When the method amethod() calls the another() method, it is sending a reference to the v (ValHold) object. The first thing another() does is reset i to 20. It is resetting the original object's value since it only passed a reference to the object, not the object itself nor a copy of the object. If you read the Campfire's "Cup Size" essay, you'll see my point. Following that logic through, you'll see that i is really 20 when it is finally printed out at the end of amethod().
[This message has been edited by Rebecca Pickett (edited April 10, 2001).]

[John M. Gabriele]

cleaned up a bit:

class ValHold
{
    public int i = 10 ; 
}
//-----------------------------------------
public class ObParm
{
    public static void main( String[] args )
    {
        ObParm o = new ObParm();
        o.aMethod();
    }
 
    public void aMethod()
    {
        int i = 99 ;
        ValHold v = new ValHold();
        v.i = 30 ;
 
        another( v , i );
 
        System.out.println( v.i );
    }
 
    public void another( ValHold v, int i )
    {
 
        i = 0 ;
        v.i = 20 ;
 
        ValHold vh = new ValHold();
 
        v = vh ;
 
        System.out.println( v.i + " " + i );
    }
}

Note that over in another(), you're monkeying with a copy of v, even though
it refers to the same object that the original v refers to. when you get back
to aMethod(), you've still got that original v waitin' there still referring
to the original ValHold object.
Umm.. also note the coding conventions in the chicken coop to make your
code look spiffy like above!

[Parag Kale]

OK...i got it now.I needed to think THAT way.
About the code part, i'll remember that. Promise clean code the next time.
Thanx a bunch.
Cheers,
Parag