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From: jh@efd.lth.se (Joergen Haegg)
Newsgroups: comp.unix.wizards
Subject: forking and zombies in SYSV.
Message-ID: <1990Aug20.133107.12516@lth.se>
Date: 20 Aug 90 13:31:07 GMT
References: <1990Aug19.210351.24632@dde.dk>
Sender: newsuser@lth.se (LTH network news server)
Reply-To: jh@efd.lth.se (Joergen Haegg)
Organization: Lund Institute of Technology, Sweden
Lines: 26

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?

-- 
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Joergen Haegg				jh@efd.lth.se	postmaster@efd.lth.se
System manager @ efd			046-107492
Lund Institute of Technology		Sweden