Edit report at http://bugs.php.net/bug.php?id=51657&edit=1
ID: 51657 Updated by: degeb...@php.net Reported by: 1234ru at gmail dot com Summary: switch() construction works wrong with empty arrays -Status: Open +Status: Bogus Type: Bug Package: Scripting Engine problem PHP Version: 5.2.13 New Comment: 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. While perhaps not immediately clear, this is correct behavior. An empty array evaluates to false and in your snippet, isset($a['a']) evaluates to false. In other words, $a == isset($a['a']), and so the first case in your switch will be used. Previous Comments: ------------------------------------------------------------------------ [2010-04-25 00:11:41] 1234ru at gmail dot com Description: ------------ When building switch() structure using arrays, if array is empty, first case always fires (while it should not) (version 5.2.12) Test script: --------------- $a = array(); switch($a) { case (isset($a['a']) ): echo 'If $a is empty, you should not see it'; break; default: echo 'Hello.'; break; } ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/bug.php?id=51657&edit=1
