PHP: Return a reference to an array element

Question

Is there a way in PHP to return a reference to an element in array?

function ref(&$array, &$ref) { $ref = $array[1]; }
$array = array(00, 11, 22, 33, 44, 55, 66, 77, 88, 99);
ref($array, $ref);
$ref = 'xxxxxxxxxx';
var_dump($ref);
var_dump($array);

I expect that $array will be changed as in the following code:

$array = array(00, 11, 22, 33, 44, 55, 66, 77, 88, 99);
$ref = &$array[1];
$ref = 'xxxxxxxxxx';
var_dump($ref);
var_dump($array);

Or read this quite good SO answer if you are not a friend of manuals stackoverflow.com/questions/2030906/… — Andresch Serj
– Andresch Serj, Commented Apr 10, 2014 at 10:10
See this documentation: php.net/manual/en/language.references.arent.php — Barmar
– Barmar, Commented Apr 10, 2014 at 10:11

vbarbarosh · Accepted Answer · 2014-04-10 11:11:14Z

27

I've found two ways to return reference to an array element:

1. Using return by reference and =&

    function & ref(&$array)
    {
            return $array[1];
    }

    $array = array(00, 11, 22, 33, 44, 55, 66, 77, 88, 99);
    $ref =& ref($array);
    $ref = 'xxxxxxxxx';
    var_dump($ref);
    var_dump($array);

2. Put reference into an array

    function ref(&$array, &$ref = array())
    {
            $ref = array();
            $ref[] = &$array[1];
    }

    $array = array(00, 11, 22, 33, 44, 55, 66, 77, 88, 99);
    ref($array, $ref);
    $ref[0] = 'xxxxxxxxx';
    var_dump($ref);
    var_dump($array);

answered Apr 10, 2014 at 11:11

vbarbarosh

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1 Comment

David Spector Jan 23 at 14:35

I chose the first method, and it worked. PHP is so strange that it passes strings by reference but arrays more or less by value (among many other peculiarities).

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