# Passing by Reference

Domains:

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);. And as of PHP 5.4.0, call-time pass-by-reference was removed, so using it will raise a fatal error.

The following things can be passed by reference:

• Variables, i.e. foo($a) • References returned from functions, i.e.:  <?php function foo(&$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 $a; } foo(bar()); // Produces fatal error as of PHP 5.0.5, strict standards notice // as of PHP 5.1.1, and notice as of PHP 7.0.0 foo($a = 5); // Expression, not variable
foo(5); // Produces fatal error

class Foobar
{
}

foo(new Foobar()) // Produces a notice as of PHP 7.0.7
// Notice: Only variables should be passed by reference
?>

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