Edit report at http://bugs.php.net/bug.php?id=52194&edit=1
ID: 52194 Updated by: [email protected] Reported by: bubi1 at mailinator dot com Summary: incorrect behaviour when printing non existend array index Status: Bogus Type: Bug Package: *General Issues Operating System: all PHP Version: 5.3.2 New Comment: Because non-integer keys are converted to integers. Everything is explained here: http://www.php.net/manual/en/language.types.string.php Previous Comments: ------------------------------------------------------------------------ [2010-06-26 19:44:45] bubi1 at mailinator dot com Yes, i know that a string is an array of chars. Maybe i'm wrong,but i remember once only the syntax $string{0} was working, non $string[0] also. Anyway, why this works? php -r '$arr = 'test';var_dump($arr['rasmus']);' still prints "t". There is no key called "rasmus" in the $arr string... ------------------------------------------------------------------------ [2010-06-26 19:12:01] [email protected] A string is an array of characters. You have: $arr[0] = "test"; Therefore array index 0 on that string is: $arr[0][0] which very logically is "t" ------------------------------------------------------------------------ [2010-06-26 18:29:04] bubi1 at mailinator dot com Description: ------------ The code below is self explaining. Test script: --------------- php -r '$arr = array('test');var_dump($arr[0][0]);var_dump(isset($arr[0][0]));' Expected result: ---------------- NULL bool(false) Actual result: -------------- string(1) "t" bool(true) ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/bug.php?id=52194&edit=1
