#51010 [Opn]: Bad example

ID: 51010 Updated by: rquadling@php.net Reported By: chris dot dickinson at web dot de Status: Open Bug Type: Documentation problem PHP Version: Irrelevant New Comment: assert() returns the state of the test. If $divisor == 0, then assert() will return true and the trigger will be called. For this sort of test, maybe a plain if($divisor == 0) is enough. Previous Comments: ------------------------------------------------------------------------ [2010-02-11 09:14:24] chris dot dickinson at web dot de Maybe the author wanted to write something like the following? : <?php if ($divisor == 0) { trigger_error("Cannot divide by zero", E_USER_ERROR); } ?> or <?php if ( ! assert($divisor != 0)) { trigger_error("Cannot divide by zero", E_USER_ERROR); } ?> ------------------------------------------------------------------------ [2010-02-11 08:31:34] chris dot dickinson at web dot de Description: ------------ Hi, On your page http://ch2.php.net/manual/en/function.trigger-error.php you give only this example <?php if (assert($divisor == 0)) { trigger_error("Cannot divide by zero", E_USER_ERROR); } ?> It seems badly chosen. If $divisor is 0 then the assertion succeeds, when actually you would probably not want $divisor to be 0 (!). If $divisor is anything else but 0, the assertion will fail and give you an error message. I got: "Warning: assert() [function.assert]: Assertion failed in ..." Reproduce code: --------------- --- From manual page: function.trigger-error#Examples --- ------------------------------------------------------------------------ -- Edit this bug report at http://bugs.php.net/?id=51010&edit=1