#49322 [Bgs]: Incorrect array assignment

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: Moreover, variables, which are references, when copied are not copied as reference and instead a true copy is made. Array elements should not be treated differently. Previous Comments: ------------------------------------------------------------------------ [2009-08-21 21:27:12] anshul at designgrill dot com Ok. But it is not obvious from the documentation. A cautionary statement should be included in there. ------------------------------------------------------------------------ [2009-08-21 21:09:20] colder@php.net Again, those are expected results, when you copy an array, its elements that are references will be copied as references. It has always been like that. ------------------------------------------------------------------------ [2009-08-21 17:21:28] anshul at designgrill dot com 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. ------------------------------------------------------------------------ [2009-08-21 17:08:22] colder@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); ?> ------------------------------------------------------------------------ The remainder of the comments for this report are too long. To view the rest of the comments, please view the bug report online at http://bugs.php.net/49322 -- Edit this bug report at http://bugs.php.net/?id=49322&edit=1