Object references, assignment and as function arguments

As of PHP5, objects are treated REFERENCE ! (same behaviour as C++ and Java)

1) assignment from an objects to another copies the reference, not the content !!

What is a reference ? it's a variabile/object that ONLY contains the pointer to the effective object in the memory. As I said in a previous post, the assigment instruction copy ONLY the value (pointer) of the object-pointer, it doesn't duplicate the content (members) of the class !

Let's demonstrate with an example, using the unset() function

class A { public $foo = 1; }

$a = new A;

$aRef = $a;

unset($aRef ); //delete ONLY the object-pointer !

print $a->foo; //print 1. This is the proof that $aRef is only a pointer

$aRef = $a;

unset($aRef ->foo); //through the object-pointer $c, it delete the real content !!

print $a->foo; //Notice: Undefined property !! This is the proof that $aRef is not a copy !!

2) objects are passed to functions by REFERENCE !

class A { public $foo = 1; }

$a = new A;

function foo($obj) {

$obj->foo = 2;

unset($obj); // delete local reference (that is a copy), no problem !

}

foo($a);

echo $a->foo; //2 ! this is the proof that the argument is passed as reference (the function doesn't return any value !) and not as a copy (the value outside the function is changed!)

Note 1: arrays are passed by copy, unless the function declare the argument as reference with "&"

function arrayDeleteFirst(&$ar) { unset($ar[0]); }

Note2: references work with variables as well

$a = 1;

$b = &$a;

$b = 2;

unset($b); //no problem. It deletes the reference, not the variabile pointed !

print $a; //2