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Passing by Reference

You can pass variable to function by reference, so that function could modify its arguments. The syntax is as follows:

<?php
function foo(&$var)
{
    
$var++;
}

$a=5;
foo($a);
// $a is 6 here
?>

Note that there's no reference sign on function call - only on function definition. Function definition alone is enough to correctly pass the argument by reference. In recent versions of PHP you will get a warning saying that "Call-time pass-by-reference" is deprecated when you use a & in foo(&$a);.

Following things can be passed by reference:

  • Variable, i.e. foo($a)

  • New statement, i.e. foo(new foobar())

  • Reference, returned from a function, i.e.:

    <?php
    function &bar()
    {
        
    $a = 5;
        return
    $a;
    }
    foo(bar());
    ?>

    See also explanations about returning by reference.

Any other expression should not be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:

<?php
function bar() // Note the missing &
{
    
$a = 5;
    return
$a;
}
foo(bar());

foo($a = 5); // Expression, not variable
foo(5); // Constant, not variable
?>

These requirements are for PHP 4.0.4 and later.

Copyright © 1997 - 2007 by the PHP Documentation Group. This material may be distributed only subject to the terms and conditions set forth in the Open Publication License, v1.0 or later. A copy of the Open Publication License is distributed with this manual, the latest version is presently available at http://www.opencontent.org/openpub/. Please see full copyright text at http://www.php.net/manual/en/copyright.php Original version of the above documentation is available at http://www.php.net/manual/en/



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