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:
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); ?> Previous Comments: ------------------------------------------------------------------------ [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