how do I make a process release its terminal?

[m5d at bobkat.UUCP]

In article <453 at vixie.UUCP> paul at vixie.UUCP (Paul Vixie Esq) writes:
>I have a program that forks itself and exits upon running -- its child
>process runs as a daemon forever after.  In 'ps aux', I notice that it
>still shows my tty as its control terminal -- even if I log out and
>back in.
>
>I see that syslogd, inetd, cron, and the rest do NOT show 'co' as their
>control terminal; how do I make this happen for me?
>
>This is (more or less) BSD 4.2, and the program is setuid root.  Thanks...
>-- 
>Paul A. Vixie        {ptsfa, qantel, crash, winfree}!vixie!paul
>329 Noe Street       dual!ptsfa!vixie!paul at ucbvax.Berkeley.EDU
>San Francisco        nike!ptsfa!vixie!paul at seismo.CSS.GOV
>CA  94116            paul at vixie.UUCP     (415) 864-7013

My guess is that the other daemons you mentioned are started in the
"rc" command file.  I think that "init" runs this with no standard file
descriptors, or maybe with /dev/null.  So far as I know, you can't get
rid of the control terminal.  You might try fiddling with the process
group, although the shell will grab the tty back anyway.

All this makes me wonder.  I don't have source; all the stuff I know
I've figured out through conjecture and the documentation.  If I ask a
question, and somebody who has source looks up the answer and posts it,
will the Unix police come and gun us all down?

--
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Mike McNally                                    Digital Lynx Inc.
Software (not hardware) Person                  Dallas  TX  75243
uucp: {texsun,killer,infotel}!pollux!bobkat!m5d (214) 238-7474