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Joseph Lee |
at Jun 17, 2004 at 11:22 am
|
⇧ | 
| |
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
Sent: Thursday, June 17, 2004 7:34 AM
To: internals@lists.php.net
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 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
tony2001@phpclub.net || antony@dovgal.com
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