Question

How does Java deal with passing reference data type arguments?? Can somebody give a clear picture?

Answer 1

Java passes a copy of the reference to the method. The reference still points to the same instance.

Imagine if I had a slip of paper with a restaurant’s address on it. You also want to go to the same restaurant so I get a new slip of paper and copy the address of the restaurant on to that paper and give it to you. Both slips of paper point to the same restaurant but they are separate references to the instance.

The restaurant itself is not duplicated, only the reference to it is duplicated.

Jon Skeet provides a similar analogy:

The balloon analogy

I imagine every
object as a helium balloon, every
reference as a piece of string, and
every variable as something which can
hold onto a piece of string. If the
reference is a null reference, that’s
like having a piece of string without
anything attached to the end. If it’s
a reference to a genuine object, it’s
a piece of string tied onto the
balloon representing that object. When
a reference is copied (either for
variable assignment or as part of a
method call) it’s as if another piece
of string is created attached to
whatever the first piece of string is
attached to. The actual piece of
string the variable (if any) is
holding onto doesn’t go anywhere –
it’s only copied.

Here is an example:

// Here I have one instance and one reference pointing to it
Object o = new Object();
// At this moment a copy of "o" is made and passed to "foo"
foo(o);

void foo(Object obj) {
    // In here I have obj which is a copy of whatever
    // reference was passed to me
}