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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:


&#60;?php
function foo(&#38;$var)
{
    $var++;
}

$a=5;
foo($a);
// $a is 6 here
?&#62;

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);.

The 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:


&#60;?php
function bar() // Note the missing &#38;
{
    $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
?&#62;

These requirements are for PHP 4.0.4 and later.

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