Re: why does exit() print its argument?

Why does exit still set the exit status then? If it was just supposed to display a message, I wouldn't think that it would manipulate the exit status. -Jason ----- Original Message ----- From: "Zeev Suraski" <zeev@zend.com> To: "Jim Winstead" <jimw@apache.org> Cc: <php-dev@lists.php.net> Sent: Friday, September 28, 2001 2:54 PM Subject: Re: [PHP-DEV] why does exit() print its argument? > That's the way it's been since forever. The documentation is wrong - exit > and die are identical. > It came up numerous times, we can have a separate shell_exit or something > along these lines in the CGI module. > > Zeev > > At 20:48 28-09-01, Jim Winstead wrote: > >in Zend/zend_execute.c, line 2387, why are we printing the value > >passed to exit? writing scripts in php for use outside of a web > >server, it would be nice to be able to set the exit value without it > >getting printed. > > > >jim > > > >-- > >PHP Development Mailing List <http://www.php.net/> > >To unsubscribe, e-mail: php-dev-unsubscribe@lists.php.net > >For additional commands, e-mail: php-dev-help@lists.php.net > >To contact the list administrators, e-mail: php-list-admin@lists.php.net > > > -- > PHP Development Mailing List <http://www.php.net/> > To unsubscribe, e-mail: php-dev-unsubscribe@lists.php.net > For additional commands, e-mail: php-dev-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >