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Passing by Reference

You can pass a variable by reference to a function so the function can modify the variable. The syntax is as follows:

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

$a=5;
foo($a);
// $a is 6 here
?>

Note: There is no reference sign on a function call - only on function definitions. Function definitions alone are enough to correctly pass the argument by reference. 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);.

The following things can be passed by reference:

Variables, i.e. foo($a)
New statements, i.e. foo(new foobar())
References returned from functions, i.e.:

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

See more about returning by reference.

No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:

 <?php
function foo(&$var)
{
    $var++;
}
function bar() // Note the missing &
{
    $a = 5;
    return $a;
}
foo(bar()); // Produces fatal error since PHP 5.0.5

foo($a = 5); // Expression, not variable
foo(5); // Produces fatal error
?>

These requirements are for PHP 4.0.4 and later.