Re: about referencing A from B
You're exactly right - to make a reference, you'll need to either 1) make A
global, or 2) pass A as a parameter to B's constructor.
Re: (un)serialising maintaining reference, not copy
Sorry, haven't tried serialising an object with references in it before - I
guess the easiest way to find out is to do a simple test. Something like (in
pseudocode)
1. define classes A and B
2. instantiate A and B
3. serialise A and B
4. unset() objects A and B
5. unserialise A and B
6. change something in A
7. see if changes are shown in B
-----Original Message-----
From: Alok K. Dhir [mailto:[EMAIL PROTECTED]]
Sent: Thursday, April 11, 2002 6:52 AM
To: [EMAIL PROTECTED]
Subject: [PHP] OO-PHP Style question
(sent this to the wrong list the first time - apologies in advance)
Given an app with the following overall class structure:
Base.class {
function Base() {
##initialize stuff
}
}
A.class extends Base {
var $var;
var $b;
function A() {
$var="x";
$b=new B.class();
}
}
B.class extends Base {
var $y;
function B() {
$y="something";
##How do I refer to $var in A
}
}
Now, how would/could an instance of B get to a variable in A (i.e. the
calling class). One (seemingly rather ugly) way is to declare the A
instance global and refer to that from within B. I.e.
<?
global $x
$x=new A; ## now we can say something like "$x->var" from within B ?>
Or, another way is to pass a reference to A in with the call to B - so
we'd change the class defs for A and B to look like:
A.class extends Base {
var $var;
var $b;
function A() {
$var="x";
$b=new B.class($this); //pass a ref to myself
}
}
B.class extends Base {
var $a;
var $y;
function B(&$a) { //always take this arg as a reference
$y="something";
// now $a contains a ref to the calling object
echo $a->var;
}
}
Am I thinking about this correctly? Am I missing some obvious language
contruct which would obviate the need for A to pass itself as an arg to
the instantiation of B? Is my object structure completely naive?
In the second case, what happens after you serialize and unserialize an
A object as in a session? Will The instance of B called $b in the A
object still contain a _reference_ to the A object in it's $a var, or
will it now have a _copy_?
I'm trying to get an idea of best practices for these types of issues.
Any comments/discussion is very much appreciated.
Al
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