Does the use keyword in PHP closures pass by reference?

Question

For example, if I do this:

function bar(&$var)
{
    $foo = function() use ($var)
    {
        $var++;
    };
    $foo();
}

$my_var = 0;
bar($my_var);

Will $my_var be modified? If not, how do I get this to work without adding a parameter to $foo?

Answer 1

No, they are not passed by reference - the use follows a similar notation like the function's parameters.

As written you achieve that by defining the use as pass-by-reference:

    $foo = function() use (&$var) 

It's also possible to create recursion this way:

$func = NULL; $func = function () use (&$func) {     $func(); } 

---

NOTE: The following old excerpt of the answer (Jun 2012) was written for PHP < 7.0. As since 7.0 (Dec 2015) the semantics of debug_zval_dump() changed (different zval handling) the refcount(?) output of it differs nowadays and are not that much saying any longer (integers don't have a refcount any longer).

Validation via the output by not displaying $my_var changed (from 0) still works though (behaviour).

You can validate that on your own with the help of the debug_zval_dump function (Demo):

function bar(&$var) {     $foo = function() use ($var)     {         debug_zval_dump($var);         $var++;     };     $foo(); };      $my_var = 0; bar($my_var); echo $my_var; 

Output:

long(0) refcount(3) 0 

A full-through-all-scopes-working reference would have a refcount of 1.

Answer 2

Closures are, almost by definition, closed by value, not by reference. You may "use by reference" by adding an & in the argument list:

function() use (&$var) 

This can be seen in example 3 in the anonymous functions manual page.

Answer 3

No, they are not passed by reference.

function foo(&$var)
{
    $foo = function() use ($var)
    {
        $var++;
    };
    $foo();
}

$my_var = 0;
foo($my_var);
echo $my_var; // displays 0

function bar(&$var)
{
    $foo = function() use (&$var)
    {
        $var++;
    };
    $foo();
}

$my_var = 0;
bar($my_var);
echo $my_var; // displays 1