ID: 49322 User updated by: anshul at designgrill dot com Reported By: anshul at designgrill dot com Status: Bogus Bug Type: Arrays related Operating System: All PHP Version: 5.2.10 New Comment:
The example you gave behaves expectedly and actually denotes the bug is valid. I would request you to carefully go through the two code snippets I gave. Even when you copy a value by reference the array, the copy is not actually made and instead the refcount is increased of the array. PHP will defer the actual copy until any one of the array is actually modified. Now when you copy the reference to the any array item, and modify using that reference, a new copy is created and values updated. But if you keep the reference of the item in a variable before array copying, modifying this variable will not force PHP to create the actual copy of the array. Incorrect behavior at the best. Previous Comments: ------------------------------------------------------------------------ [2009-08-21 17:08:22] col...@php.net Yes, the array itself is copied by value, but not the array content, as you discovered. $a = array(); $b = $a; $b[] = 2; var_dump(count($a)); // int(0) ------------------------------------------------------------------------ [2009-08-21 17:00:12] anshul at designgrill dot com Following will give a different result. Just because reference is copying after assignment. PHP manual says the assignment operation copies by value for arrays. <?php $arr1 = array(1, 2); $arr2 = $arr1; $var = &$arr1[0]; $var = 10; print_r($arr1); print_r($arr2); ?> ------------------------------------------------------------------------ [2009-08-21 16:54:32] col...@php.net Sorry, but your problem does not imply a bug in PHP itself. For a list of more appropriate places to ask for help using PHP, please visit http://www.php.net/support.php as this bug system is not the appropriate forum for asking support questions. Due to the volume of reports we can not explain in detail here why your report is not a bug. The support channels will be able to provide an explanation for you. Thank you for your interest in PHP. That's expected, arrays have always worked like that. ------------------------------------------------------------------------ [2009-08-21 16:52:29] anshul at designgrill dot com Description: ------------ If you keep reference to an element of the array before the copy operation using = operator, you can still modify both the arrays using the reference copied earlier. The same doesn't happen if you keep the reference after the copy operation and try to modify the variable using it. Reproduce code: --------------- <?php $arr1 = array(1, 2); $var = &$arr1[0];//Keeping the reference $arr2 = $arr1;//Copy operation $var = 10;//Will modify both the arrays, bad print_r($arr1); print_r($arr2); ?> Expected result: ---------------- Array ( [0] => 10 [1] => 2 ) Array ( [0] => 1 [1] => 2 ) Actual result: -------------- Array ( [0] => 10 [1] => 2 ) Array ( [0] => 10 [1] => 2 ) ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=49322&edit=1
