Bug #1269 Updated: is_array function defines the variable?

ID: 1269 Updated by: zeev Reported By: arneodo@initiative.fr Status: Closed Bug Type: Misbehaving function Assigned To: Comments: is_array() doesn't create the variable, it is created before the call to is_array(), and passed to it. You need to understand how PHP's automatic declaration of variables work. In essence: As soon as you reference a variable, it is automatically defined. For example, if you pass a variable to is_array(), like you have in your example, PHP checks out whether it exists, if it doesn't - it creates it, and then it passes it over to is_array(). It has to pass *something*, that's why the variable is automatically defined. The implementation of is_array() has nothing to do with this behavior - each and every function in PHP would behave in the same way. There aren't too many exceptions to this case, basically, only three special functions may reference a variable without causing it to be implicitly declared - those are unset(), isset() and empty(). Full Bug description available at: http://ca.php.net/bugs.php3?id=1269 -- PHP Development Mailing List http://www.php.net/ To unsubscribe send an empty message to php-dev-unsubscribe@lists.php.net For help: php-dev-help@lists.php.net