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

?>

Зауваження: 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.

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 a notice

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

?>