I guess "exit();" terminates execution within itself without returning to the caller, so that is no chance of getting a runtime error.
For example, " return ( exit() ); " is wrong, but works. Joe Lee -----Original Message----- From: Jason Garber [mailto:[EMAIL PROTECTED] Sent: Thursday, June 17, 2004 7:34 AM To: [EMAIL PROTECTED] Subject: Re: [PHP-DEV] Throw Question That's what I figured. throw is a language construct. However, from the manual (http://php.net/exit): void exit ( int status) Note: This is not a real function, but a language construct. Why does $x || exit; work without a parse error? Thanks, Jason Garber At 6/17/2004 10:22 AM +0400, Antony Dovgal wrote: >On Thu, 17 Jun 2004 02:17:26 -0400 >Jason Garber <[EMAIL PROTECTED]> wrote: > > > Consider the following: > > Why is this? > >Take a look at the bug #28727: >http://bugs.php.net/?id=28727 > >--- >WBR, >Antony Dovgal aka tony2001 >[EMAIL PROTECTED] || [EMAIL PROTECTED] > >-- >PHP Internals - PHP Runtime Development Mailing List >To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Internals - PHP Runtime Development Mailing List To unsubscribe, visit: http://www.php.net/unsub.php
