Java Fundamental

Question

Hi!,
I have one very basic question.
Java passes all parameters except literals by reference. right!
So why doesn't the change in a function to one such parameter after its value when I print it outside the function?
eg.
I have a function copyElement(DefaultMutableTreeNode node1, DTMN node2)
node2=(DTMN)node1.clone();
and the place where I call it
I first create
DTMN node1=new DTMN("NODE1");
DTMN node2=new DTMN("NODE2");
copy(node1,node2)
then when I print node2, it still shows me NODE2
This is because clone returns a new ref. pointer, which will be lost once it comes out of the function.
So is there a way by which I can create an object outside the function,a dn in the function assign it a value.
Or am I totally on the wrong track???
thanks

Answer 1

Perhaps it would be clearer to you if you wrote it like this:public void copyElement( Object nodeX, Object nodeY )
    nodeY = nodeX.clone();
public void test()
    Object node1 = new String( "NODE1" );
    Object node2 = new String( "NODE2" );
    copy( node1, node2 );
}Changing what the variable nodeY points in your copyElement method does not impact what the variable node2 points to. You are reassinging the object that nodeY referes to but this does not change the object that node2 refers to. You are not altering the actual object that is being passed to copy (although you could), rather you are just reassigning the variable reference nodeY, but the scope of that variable is only within the copy method.
Perhaps a better way to write your method would be:public Object copyElement( Object nodeX )
    Object nodeY = nodeX.clone();
    return nodeY;
public void test()
    Object node1 = new String( "NODE1" );
    Object node2 = copyElement( node1 );
}