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?
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3 Answers
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) therefcount(?)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
$my_varchanged (from0) still works though (behaviour).
You can validate that on your own with the help of the debug_zval_dump function (Demo):
function bar(&$var) { $foo = function() use ($var) { debug_zval_dump($var); $var++; }; $foo(); }; $my_var = 0; bar($my_var); echo $my_var; Output:
long(0) refcount(3) 0 A full-through-all-scopes-working reference would have a refcount of 1.
1 Comment
elite5472
Better a year later than never!
Closures are, almost by definition, closed by value, not by reference. You may "use by reference" by adding an & in the argument list:
function() use (&$var) This can be seen in example 3 in the anonymous functions manual page.
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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 