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
On Thu, 17 Jun 2004 02:17:26 -0400
Jason Garber jason@ionzoft.com 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
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 jason@ionzoft.com 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