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PHP

Passing by Reference


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.



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