Exactly, but the proper "Computer Science" terminology for what you call
"Pass-By-Value in its passing of object references" is actually
"Pass-By-Reference".
(*Chris*)
----- Original Message -----
From: Ted Neward <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, August 13, 1999 4:21 AM
Subject: Re: will this code work?
> Chris, I don't get it--what are you proving here? That I can call
obj.add()
> and get 10 back? That's precisely what I would expect, and doesn't
disprove
> what I've claimed earlier.
>
> Look, I think we're getting bogged down in semantics here. When I say that
> Java uses Pass-By-Value in its passing of object references, what I mean
is
> that the *pointer* (the Java reference itself, not the object) is copied,
> and the object remains "out there somewhere". I'm not, in any way,
claiming
> that the entire Object gets copied onto the stack, and any method call on
> that object results in a modification of state of that temporary Object
> only.
>
> In my earlier example:
>
> public class Param
> {
> public static void main(String[] args)
> {
> String temp = null;
> manipParameter(temp);
> System.out.println("temp = " + temp);
> }
>
> private static void manipParameter(String param)
> {
> param = "This should be modified";
> }
> }
>
> Consider this (crude) pictogram of what's happening in memory. Just before
> the call to manipParameter(), the references in question look like this:
>
> temp ---> (Nothing: for purposes of argument, assume address 0x00000000)
>
> Now, we make the call to manipParameter; a new "pointer" to a String type
is
> created, pushed on the stack, and control jumps to the start of the
> manipParameter method. This new "pointer" is called param within the body
of
> manipParameter:
>
> temp ---> (0x00000000)
> param ---> (whatever temp was pointing to, in this case 0x00000000)
>
> When we execute param = "This should be modified", it looks like this:
>
> temp ---> (0x00000000)
> param ---> String @ 0x12345678 ("This should be modified")
>
> But because we don't modify the actual "temp" reference itself, temp still
> retains its reference pointing to null when it returns!
>
> I really think we're getting caught up in a semantic difference here, and
> that if we were to set out our own "Glossaries of terms", we'd find we're
> saying the same thing using different terms. I'm a C++-head, originally,
so
> I can see this in terms of passing the pointers-by-value.
>
> What's more, you're sort of proving my point--you say "you're changing the
> value of the reference, not the value of the String", and that's precisely
> what I'm trying to prove. You can change the value of the reference, which
> won't, in turn, affect the value of the reference used to create the
> parameter. Otherwise, if pointing "param" to a new String also affected
> "temp" (the source of the parameter to the method), then "temp" would have
> the reference to modified contents when we returned out, and it obviously
> doesn't.
>
> (And just as a source of nit-picking, I think the second line in test.main
> should read process_me(obj), correct?)
>
> If you want to get truly technical about it, Java uses Pass-by-Value for
ALL
> of its types, including Objects; but because Java only knows about
> heap-allocated objects, passing an object "by value" involves passing the
> pointer to that Object by value. It's a subtle point, one lost on many
Java
> developers, but once it's understood, it tends to make a LOT of things a
lot
> clearer.
>
> Is this any clearer?
>
> Ted Neward
> Patterns/C++/Java/CORBA/EJB/COM-DCOM spoken here
> http://www.javageeks.com/~tneward
> "I don't even speak for myself; my wife won't let me." --Me
>
> -----Original Message-----
> From: Chris Pratt <[EMAIL PROTECTED]>
> To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
> Date: Tuesday, August 10, 1999 11:14 PM
> Subject: Re: will this code work?
>
>
> >Wrong, Wrong, Wrong. Your example is correct, but that's because you're
> >changing the value of the reference, not the value of the String
(remember
> >String's are immutable). Java uses Pass by Reference for Objects and
Pass
> >by Value for intrinsic types (int, float, char, etc). This has been a
> major
> >source of confusion with the Java detractors saying that the entire
> contents
> >of each Object must be copied to the stack since Java is "pointerless".
If
> >you don't believe me, try this program. I bet you get "i = 10"
> > (*Chris*)
> >
> >class Obj {
> > int i;
> >
> > public Obj (int i) {
> > this.i = i;
> > }
> >
> > public int add (int j) {
> > i += j;
> > return i;
> > }
> >
> > public String toString () {
> > return String.valueOf(i);
> > }
> >
> >}
> >
> >public class test {
> >
> > public static void process_me(Obj o) {
> > o.add(5);
> > }
> >
> > public static void main(String[] args) {
> > Obj obj = new Obj(5);
> > obj.add(5);
> > System.out.println("obj = " + obj);
> > }
> >
> >}
> >
>
>
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