#30292 [Bgs]: Undefined array variable gives a problem with the output

ID: 30292 User updated by: devish at zonnet dot nl Reported By: devish at zonnet dot nl Status: Bogus Bug Type: Output Control Operating System: Windows XP PHP Version: 5.0.1 New Comment: I was first thinking of that to, but that wasn't it. A qoute out of my php.ini error_reporting = E_ALL ; Print out errors (as a part of the output). For production web sites, ; you're strongly encouraged to turn this feature off, and use error logging ; instead (see below). Keeping display_errors enabled on a production web site ; may reveal security information to end users, such as file paths on your Web ; server, your database schema or other information. display_errors = On Previous Comments: ------------------------------------------------------------------------ [2004-09-30 15:16:13] tony2001@php.net You will not get a notice here, but a parse error, because this syntax is not valid. Set display_errors to "on" and you should see it. ------------------------------------------------------------------------ [2004-09-30 15:03:59] devish at zonnet dot nl Description: ------------ When declaring an array but don't initialize it, it stops running the script. When declaring a normal variable it will just produce a Undefined variable notice. In the reproduce code it will not print "hello" Offcourse this sort of code is rearly used, but the reason for reporting this is that it doesn't report a notice or any warning. Im running Win XP, with Apache 2.0 and only have one extension in my php.ini (the mysql extension) Reproduce code: --------------- <?php error_reporting(E_ALL); $int[]; print "hello"; ?> Expected result: ---------------- I expect: Notice: Undefined variable: int in E:\www-root\sqlite\test.php on line 4 hello I See: white screen ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=30292&edit=1