Edit report at http://bugs.php.net/bug.php?id=50230&edit=1

 ID:                 50230
 Updated by:         ar...@php.net
 Reported by:        ninzya at inbox dot lv
 Summary:            References passed to closures as variables corrupt
                     original passed variable
-Status:             Verified
+Status:             Closed
 Type:               Bug
 Package:            Scripting Engine problem
 Operating System:   *
 PHP Version:        5.3, 6
-Assigned To:        
+Assigned To:        arpad
 Block user comment: N
 Private report:     N

 New Comment:

Thank you for your bug report. This issue has already been fixed
in the latest released version of PHP, which you can download at 
http://www.php.net/downloads.php

Fixed by r308320 which is in 5.3.6


Previous Comments:
------------------------------------------------------------------------
[2010-05-28 21:52:03] dchurch at sciencelogic dot com

I feel that this is probably the same problem:



$array = array(1,4,2,3);

usort($array, function($a,$b) use ($array) {

  return $a > $b ? 1 : -1;

});



Result is that $array is not changed by the usort function.

------------------------------------------------------------------------
[2009-12-10 08:01:49] ar...@php.net

The problem is the data pointer from the hash apply being overwritten in 
zval_copy_static_var() (Zend/zend_closures.c).



The following patch works for me:



http://spellign.com/patches/php-closure-ref-HEAD.patch

------------------------------------------------------------------------
[2009-11-19 13:10:49] ninzya at inbox dot lv

Description:
------------
See the reproduce code. I have a variable $ref, which is a reference to another 
variable. I am passing $ref to closure as use() by value, but, this kind of 
passing corrupts $ref variable after definition of closure and $ref becomes no 
longer reference, but points directly to copied data of $source variable, which 
$ref was previously referring to. However, if i define closure the following 
way:



$closure =function() use( &$ref) {// note pass-by-reference

  echo $ref;

};



the $ref does not loose it's state.

Reproduce code:
---------------
$source ='Dmitry';

$ref =&$source;



$closure =function() use( $ref) {

  echo $ref;

};



$ref ='Dmitry2';



echo $ref ."\n";

echo $source ."\n";

Expected result:
----------------
Dmitry2

Dmitry2

Actual result:
--------------
Dmitry2

Dmitry


------------------------------------------------------------------------



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