forking and zombies in SYSV.
Joergen Haegg
jh at efd.lth.se
Mon Aug 20 23:31:07 AEST 1990
A question about how to fork() in SYSV.
I want to fork, and the parentprocess should continue without
doing a wait().
When the child exits, it is transformed into a zombieprocess.
The zombie exists until the parent exits.
This is as it should be according to SVID.
How do I avoid getting this zombie?
The only way, am I told, is to let the child fork again and exit.
The parent will wait() for the first child, but not the child of
the first child.
The second child will not create a zombie, because the parent
(the first child) is dead.
This seems a bit silly, because /etc/init forks all the time and without
any zombies. (Ok, init is maybe a little special :-)
Is there not a cleaner way to do this?
--
--
Joergen Haegg jh at efd.lth.se postmaster at efd.lth.se
System manager @ efd 046-107492
Lund Institute of Technology Sweden
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