ID: 39992
Updated by: [email protected]
Reported By: [email protected]
-Status: Feedback
+Status: Open
Bug Type: Program Execution
Operating System: linux
PHP Version: 6CVS-2006-12-30 (CVS)
New Comment:
Yes, I'm still able to reproduce it. In some systems you may need
higher proviledges in order to reproduce this bug.
Previous Comments:
------------------------------------------------------------------------
[2008-08-12 16:28:58] [email protected]
Nuno, is this still valid bug? I can't reproduce it even with PHP
5.2.6..
------------------------------------------------------------------------
[2006-12-30 15:54:27] [email protected]
Ah I forgot: removing the "2>&1" part everything works fine.
------------------------------------------------------------------------
[2006-12-30 15:52:20] [email protected]
Description:
------------
With the short reproduce code below, PHP fork()s a new process (sh),
that itself forks a new one (php). proc_terminate() kill()s the sh
process, but the php one doesn't get killed.
I'm not really sure what is causing this problem, but it is breaking
run-tests.php on timeout.
Reproduce code:
---------------
<?php
$cmd='php -r "while(1){}" 2>&1';
$proc = proc_open($cmd, array(), $pipes);
var_dump(proc_terminate($proc));
?>
------------------------------------------------------------------------
--
Edit this bug report at http://bugs.php.net/?id=39992&edit=1
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