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);
// $a is 6 here
?>
Note: There is no reference sign on a function call - only on function definitions. Function definitions alone are enough to correctly pass the argument by reference.
The following things can be passed by reference:
-
Variables, i.e.
foo($a) -
References returned from functions, i.e.:
See more about returning by reference.<?php
function foo(&$var)
{
$var++;
}
function &bar()
{
$a = 5;
return $a;
}
foo(bar());
?>
No other expressions should be passed by reference, as the result is undefined. For example, the following examples of passing by reference are invalid:
<?php
function foo(&$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
?>
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