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Bug #17432 PHP ignores a exit-command
Submitted: 2002-05-26 03:34 UTC Modified: 2002-05-26 05:16 UTC
From: e05 at slashdot dot ch Assigned:
Status: Not a bug Package: Scripting Engine problem
PHP Version: 4.2.1 OS: Microsoft Windows 2000 SP2
Private report: No CVE-ID: None
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From: e05 at slashdot dot ch
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 [2002-05-26 03:34 UTC] e05 at slashdot dot ch 
I wrote a script with an error, and to ignore it I write an exit-command before. But PHP ignores it.

A script that shows this bug:
<?php
phpinfo();
exit(1);
if(
phpinfo();
?>
--
My configuration:
Apache 1.3.24 (win32-build from httpd.apache.org)
PHP 4.2.1 (offical win32 build)
PHP runs as Apache module.
--

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 [2002-05-26 04:09 UTC] derick@php.net 
A SCript is parse before it is executed. If it can not be parsed correctly due to a parse error, it will not get executed at all, and thus your exit(1) call has no effect here.

Not a bug -> bogus
 [2002-05-26 05:09 UTC] e05 at slashdot dot ch 
I think it?s illogical.
Why?
If you call an function which doesn?t exists, it works.
 [2002-05-26 05:12 UTC] correction at nospam dot please 
I think it?s illogical.
Why?
If you call an function which doesn?t exists, it works.
 [2002-05-26 05:15 UTC] derick@php.net 
A undefined functionis not a _parse_ error. Function names are evaluated at runtime in the Zend engine (well, uh, sometimes :), so it's quite logical from a techincal point of view.

Derick
 [2002-05-26 05:16 UTC] mfischer@php.net 
It's not "illogical". It's simply the way it works and won't change, you'll have to live with it.

Scripts are completely parsed *before* they're executed. That's a fact. If the script can't be parsed, it can't be executed.

Calling missing function is something else, it's a run-time behaviour, that's why this works.
 
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