Re: why does exit() print its argument?
| From: | Zeev Suraski | Date: | Sat, 29 Sep 2001 00:12:51 +0000 |
| Subject: | Re: why does exit() print its argument? | ||
| References: | 1 2 3 | Groups: | php.dev |
| Request: | Send a blank email to php-dev+get-66808@lists.php.net to get a copy of this message | ||
At 23:27 28-09-01, Jim Winstead wrote:
On Fri, Sep 28, 2001 at 11:14:04PM +0200, Zeev Suraski wrote: At 22:17 28-09-01, Jason Greene wrote:Uhm, I'm not dumb :) Of course it enables you to do that, but that's the wrong place to put it. exit() has, since its introduction, been a complete equivalent of die(), and had nothing to do with exit statuses. Putting this code there was wrong (even though who knows, maybe I even did that). ZeevWhy does exit still set the exit status then?Ok, apparently it does (didn't recall that it does). I'm not sure what the logic behind this is. so you can have a script that does an exit(1) and things calling that script can know that there was some sort of error in the script execution.