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.