Path: utzoo!utgpu!jarvis.csri.toronto.edu!cs.utexas.edu!usc!zaphod.mps.ohio-state.edu!think!snorkelwacker!mit-eddie!bu.edu!buengc!bph
From: bph@buengc.BU.EDU (Blair P. Houghton)
Newsgroups: comp.unix.questions
Subject: Re: Killing the correct process
Message-ID: <5382@buengc.BU.EDU>
Date: 15 Feb 90 20:28:09 GMT
References: <22332@adm.BRL.MIL> <5312@star.cs.vu.nl> <1212@root44.co.uk>
Reply-To: bph@buengc.bu.edu (Blair P. Houghton)
Followup-To: comp.unix.questions
Organization: Boston Univ. Col. of Eng.
Lines: 54

In article <1212@root44.co.uk> gwc@root.co.uk (Geoff Clare) writes:
>In article <5312@star.cs.vu.nl> maart@cs.vu.nl (Maarten Litmaath) writes:
>>In article <22332@adm.BRL.MIL> marwood@ncs.dnd.ca (Gordon Marwood)
>>>[wants to timeout ftp.]
>>
>>How about using the following general purpose script?  If that's out of the
>>question, the script might still give you a hint how to solve your problem.
>
>[ extremely complicated script deleted ]
>
>There is a much simpler way than Maarten's.

Edited for brevity...

Called with "foo (time in seconds) (command and arguments)"

>#! /bin/sh
>pid=$$
>(
>	sleep "$time"
>	# use SIGTERM first to allow process to clean up
>	kill $pid >/dev/null 2>&1

At this point, and I'll admit it's a rare possibility, but
not an impossibility, especially on multiprocessor machines,
what happens if the $command (see below) has already exited
and some other process (possibly on another processor) has 
begun with the same pid?  Or won't that happen while this
backgrounded process is tying up the process-group number?

>	sleep 2
>	# if process hasn't died yet, use SIGKILL
>	kill -$SIGKILL $pid >/dev/null 2>&1
>) &
>
>exec "$command"

I'm trying to implement exactly this control structure in C, now,
and one of my least favorite problems is that of getting
the status of a process I may not own without having to do

	system("ps -l##### > /tmp/foo"); /* ##### is the pid */

(or fork-and-exec, etc...  Actually, the system() call
wouldn't be much of a diseconomy, because it only has to
execute the ps(1) once or twice after what may be several
hours of sleep()'ing.)

I mean, just how the heck does ps(1) do it?  Everything I've seen
implies it goes through /dev/mem one byte at a time.


				--Blair
				  "Oh, lovely..."