reference passing

[reubin haz]

class ValHold{
        public int i = 10;
}
 
public class ObParm{
public static void main(String argv[]){
        ObParm o = new ObParm();
        o.amethod();
        }
        public void amethod(){
                int i = 99;
                ValHold v = new ValHold();
                v.i=30;
                another(v,i);
                System.out.print( v.i );
        }//End of amethod
 
        public void another(ValHold v, int i){
                i=0;
                v.i = 20;
                ValHold vh = new ValHold();
                v =  vh;
                System.out.print(v.i);
        }//End of another
 
}
  

I don't understand why System.out.print(v.i); in the 'another' method is printing value of 10 and in the 'amethod' method is printing value of 20. Can someone please explain to me? Thanks.

[bhavesh bhanushali]

inside amethod
---------------

1] v ----> 10

v.i = 30 ;

2] v ----> 30

another(v,i) ;

3] call is made to method where the refernce of the object v and value of variable i
is passed

inside another
--------------

v.i = 20;

1] v ----> 20

the state of the object v is changed ( ie. the value of instance variable i )

ValHold vh = new ValHold();

2] vh -----> 10 ( new object vh is created )

v = vh;

3] v ----> 10 ( this is the copied reference of v and not the object v that is
inside method amethod )
vh ----> 10

now we again return to the method amethod

inside amethod
--------------

the original refernce of the object v still remains the same as was before the call
, but the state has changed because of bthe assignment that had taken place in the
method another

[vjy chin]

The value printed in another is 10.
If you see the code

 ValHold vh = new ValHold();
                v =  vh;
 

you will understand. See here we are setting the reference of the ValHold object vh to v,and since both are same type, they both now refer to the same object. So if you access the variable i, which has the value 10, the result will be printed as 10.

But in the other case, where you are passsing the object itself to another method in the code like

 another(v,i); 

the reference is passed. So when you change the value v.i=20 in another method, that value is assigned to v.i now and the value 30 is erased. Also objects are passed by reference, when you are printing the value in amethod you get 20.
Hope it helps

[reubin haz]

Originally posted by vjy chin:

the reference is passed. So when you change the value v.i=20 in another method, that value is assigned to v.i now and the value 30 is erased.
Hope it helps

This is true, but AFTER this, v is assigned to a new ValHold instance, whose variable i is 10:

ValHold vh = new ValHold();
v = vh;

So 10 is the latest value i holds, isn't it? Then why after 'another' method finishs, i is 20? Would this mean v in 'amethod' is different from v in 'another' method? We should say Java is pass-by-value for any type data, but we often explain things in pass-by-reference, this also confuses me.

I'm still haven't clear this thought. Someone please explain more. Thanks.

[Pramod Chris]

In the method another, the state of the object pointed to by
v is changed. ie., i is now 20. By doing
v = vh;
the reference v points to the object created at
ValHold vh = new ValHold();
so v.i is now 10.
Since Java passes parameters strictly by value, once back in the method amethod, v would not point to the object that vh is pointing to, anymore.
But since you had passed the object reference v into another(), where its state was changed, back in amethod it still points to the very same object whose state was changed. Hence it still prints 20.

[reubin haz]

Thanks, Pramod. I have it cleared now.