Re: why does exit() print its argument?

On Fri, 28 Sep 2001, 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: > > >Why 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. > > (personally, i'd love to see exit() stop outputting the status, > leaving that for die(), but whatever. i know better than to argue for > consistency with perl. :) Why not make exit() behaves like this: exit(2): print nothing, set result status to 2 exit("error"); print "error", set no result. I.e, on a numeric status, exit with that status, and if it's a string, just print it. This way almost everybody gets what they want, AND it would not break much scripts either. Derick --------------------------------------------------------------------- PHP: Scripting the Web - www.php.net - derick@php.net SRM: Site Resource Manager - www.vl-srm.net ---------------------------------------------------------------------