PHP: Alternative to variables passed by reference? - php

Question

In the minutes php6 developer meeting, i came across a point where it said that call-time-pass-by-reference is no longer there in PHP6. For example, following both are incorrect for PHP6:
<?php
$foo =& new StdClass();
?>
<?php
function &foo()
{
return new StdClass();
}
$f = foo();
?>
If we can't use something like this in PHP6:
$foo =& new StdClass();
What is the alternative to that, is there any way to mimic that?
EDIT:
Ans what about variables in PHP6, can we do that for variables eg:
$this->data =& $_SESSION;

Answer 1

You don't pass parameters by reference, but you still can/must declare your function/method as receiving parameters by reference.
i.e., you don't do this (passing parameters by reference) :
my_function(& $my_var);
function my_function($a) {
// ...
}
But you can do this (declaring the function as receiving parameters by reference) :
my_function($my_var);
function my_function(& $a) {
// ...
}
And... The code examples you gave are not related to call-time-pass-by-reference, but are related to return-by-reference.
For that second thing, what if you just remove the & ? The instance of the object that's been created inside the function will be returned, and you'll still be able to work with it, won't you ?

Answer 2

Objects are always passed by reference since PHP5, so this:
$foo =& new StdClass();
is the same as:
$foo = new StdClass();
.

Answer 3

Ans what about variables in PHP6, can we do that for variables eg:
$this->data =& $_SESSION;
I see no reason why not -- PHP will not be removing references, as they are far too useful. There is no other way to create a reference to a variable.
Call-time pass-by-reference has been deprecated for a while. It just means that instead of doing this:
function foo($a) { return ++$a; }
foo(&$my_a);
You would do this:
function foo(&$a) { return ++$a; }
foo($my_a);
This leads to a much cleaner and easier-to-understand programming style, and it also ensures that variables are always referenced when you expect, and not when you don't.
Dealing with objects is a special case -- they are always passed by reference, and the only way to simulate pass-by-value is to use clone:
$a = new StdClass();
$a->foo = 'bar';
$b = $a;
$a->foo = 'qux';
// $b->foo is 'qux' too
$a = new StdClass();
$a->foo = 'bar';
$b = clone $a;
$a->foo = 'qux';
// $a->foo is 'qux' but $b->foo is 'bar'
Hope that helps!