Shadowing Object References?

[Joe San]

Hi Ranchers,

The following is an example from the K&B book on page 208.

This is regarding shadowing of object references. I fail to understand why the myBar variable declared in the changeIt method receives a reference to the same bar object? What this actually implies. Does it mean that shadowing does not apply when it comes to object references?

Regards,
Jothi Shankar Kumar. S

[Barry Gaunt]

Have you not omitted something? Such as formatted code between

tags?

[Joe San]

Hi Barry,

Sorry...I have forgotten to put the code in place. So here goes below,

class Bar{
int barNum = 28;
}
class foo{
Bar myBar = new Bar();
void changeIt(Bar myBar){
myBar.barNum = 99;
System.out.println("mybar.barnum in changeit is "  + myBar.barNum);
myBar = new Bar();
myBar.barNum = 420;
System.out.println("mybar.barnum in changeit is now "  + myBar.barNum);
}
public static void main(String args[]){
Foo f = new Foo();
System.out.println("f.mybar.barnum in changeit is "  + f.myBar.barNum);
f.changeIt(f.myBar);
System.out.println("f.mybar.barnum after changeit is "  + f.myBar.barNum);
}
}

My question is why object references are not being shadowed unline primitives?

Regards,
Jothi Shankar Kumar. S

[Joe San]

Hi Ranchers,

Anyone please provide me with reason for the code above?

Regards,
Jothi Shankar Kumar. S

[wise owen]

class Bar{
   int barNum = 28;
}
class foo {
Bar myBar = new Bar();
    void changeIt(Bar myBar){ 
       [B]//The parameter myBar shadows the instance variable myBar
          //but they refer to the same object in your example.[/B]
  myBar.barNum = 99;
  System.out.println("mybar.barnum in changeit is "  + myBar.barNum);
 
  myBar = new Bar();
       [B]//The parameter myBar shadows the instance variable myBar
          //and also shows that Java always pass by value.[/B]
  myBar.barNum = 420;
  System.out.println("mybar.barnum in changeit is now "  + myBar.barNum);
}
    public static void main(String args[]){
  foo f = new foo();
  System.out.println("f.mybar.barnum in changeit is "  + f.myBar.barNum);
  f.changeIt(f.myBar);
  System.out.println("f.mybar.barnum after changeit is "  + f.myBar.barNum);
}
}

[ October 17, 2006: Message edited by: wise owen ]

[Ramu Malur S R]

Hi Jothi,

The reference myBar in changeIt(Bar myBar) is a copy of the reference f.myBar declared in main(), not the myBar member variable(which is this.myBar).

Thus the new object created is assigned to the reference copy. But the reference in main still refers to the same original object.

- Ramu
"Excellency is rarely found, more rarely valued"

[Joe San]

Im terribly confused. I did not quite understand.

[Aniket Patil]

Whenever a variable (be it a reference or a primitive) is assigned to another, what gets assigned is a value stored in the variable.

In case of primitives, this is the actual value, whereas for references, this is a "way" to get to the object on heap.

When changeIt() is called, the argument value to changeIt(), which indicates how you can get to the object, is stored in mybar (the parameter of changeIt()).

When num is set to 99, both the mybar references refer to the same object. Hence, the original object is changed.

But later on in changeIt(), mybar (parameter) is made to point to a new object. Now what is available within the scope of changeIt() is the instance member mybar pointing to the original object, and a local variable mybar pointing to a new object. Here is where shadowing takes place, and you end up using the parameter of changeIt() to refer to the new object (unless you use this.mybar).
[ October 17, 2006: Message edited by: Aniket Patil ]