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]
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