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Throw Question

php.internals

Jason Garber

22 years ago

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

Antony Dovgal

22 years ago

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

Jason Garber

22 years ago

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:

Joseph Lee

22 years ago

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:jason@ionzoft.com]
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 <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
>
>--
>PHP Internals - PHP Runtime Development Mailing List
>To unsubscribe, visit: http://www.php.net/unsub.php

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Hartmut Holzgraefe

22 years ago

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 you can use
them in expressions ...

--
Hartmut Holzgraefe  <hartmut@php.net>

Jason Garber

22 years ago

Thanks for the good explanation.

~Jason

At 6/17/2004 02:10 PM +0200, Hartmut Holzgraefe wrote: