Difference in operators for PHP

Question

What is the difference between swap($a, $a) and with an '&' added for swap(&$a, &$a)

<?php 
 function swap($x, $y) { 
 $x = $x + 1; 
 $y = $y + 2; 
 return $x * $y; 
 } 

 $a = 2; 
 $b = swap($a, $a); 
 print "$a, $b"; 
 $b = swap(&$a, &$a); 
 print "$a, $b"; 
?>

Note: not asking for the answer, just wrote this so i could provide an example. Thank You.

Sorry meant to write PHP been doing javascript work all day and it messed my brain up :P — user3364498, Apr 20 '14 at 17:51
[PHP: References Explained](http://www.php.net/manual/en/language.references.php) — h2ooooooo, Apr 20 '14 at 17:53
Thanks for that list, will help a lot. Solved my question although cannot see &$ on the list — user3364498, Apr 20 '14 at 17:54

score 1 · Answer 1 · answered Apr 20 '14 at 18:20

The & means that the parameter is passed by reference. The question is indeed a duplicate but please be aware that you shouldn't use call-time pass-by-reference anymore.

From the passing by reference page on php.net:

As of PHP 5.3.0, you will get a warning saying that "call-time pass-by-reference" is deprecated when you use & in foo(&$a);. And as of PHP 5.4.0, call-time pass-by-reference was removed, so using it will raise a fatal error.

What you still can do is using the & in the function definition:

function swap(&$x, &$y) { 
    $x = $x + 1; 
    $y = $y + 2; 
    return $x * $y; 
}

Difference in operators for PHP

1 Answers1