ID: 43538 Updated by: [EMAIL PROTECTED] Reported By: warren at transfusionmedia dot com Status: Open Bug Type: Feature/Change Request Operating System: Win XP SP2 PHP Version: 5.2.5 New Comment:
Seems odd why its coded to return 1 imo, heres a patch against latest PHP_5_3 cvs that changes this to return false: http://www.phpfi.com/332583 Previous Comments: ------------------------------------------------------------------------ [2007-12-08 20:30:47] warren at transfusionmedia dot com Description: ------------ When doing count($var) where $var is not an array, the return = 1. The problem is when you're doing a check in your code: if (count($var) > 0){ //do something } This will evaluate as true if $var is not an array. This is a problem if you are doing this: if (count($var) > 0){ sort($var); // anything that requires $var to be an array } If there is anything in that if statement that requires $var to be an array, there will be a fatal error. I know you can also add an is_array check in there, but that is non-obvious and shouldn't be necessary. Really, if $var is not an array, count($var) should return false or null, or at least a -1 Reproduce code: --------------- $var = "foo"; if (count($var) > 0){ sort($var); } else { echo "The value of var is not greater than zero"; } Expected result: ---------------- The value of var is not greater than zero Actual result: -------------- Warning: sort() expects parameter 1 to be array, string given ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=43538&edit=1
