Thanks for the good explanation.
~Jason
At 6/17/2004 02:10 PM +0200, Hartmut Holzgraefe wrote:
Joseph Lee wrote:
I guess "exit();" terminates execution within itself without returning
to the caller, so that is no chance of getting a runtime error.
parse error != runtime error
but language construct
Joseph Lee wrote:
I guess "exit();" terminates execution within itself without returning
to the caller, so that is no chance of getting a runtime error.
parse error != runtime error
but language constructs like exit, unset and print are especialy
ment to be as function-like as possible, thats why y
TED]
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 const
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, An
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
Consider the following:
$x = FALSE;
$x || throw new exception('Some Assertion');
I get the following Parse Error:
error: parse error, unexpected T_THROW in /home/.../Z_Record.php on line 153
However, this code produces no errors..
$x = FALSE;
$x || exit;
Why is this?
Thanks,
Jason Garber
--
PHP Int
