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Top Forums UNIX for Dummies Questions & Answers starting processes with timeout? Post 302072791 by Perderabo on Sunday 7th of May 2006 11:58:44 AM
Old 05-07-2006
This is an odd thing to do. But I just got this script to work:
Code:
#! /usr/bin/ksh

sleep 200 ; kill -term -$$ &
sleep 7
kill -term -$$

One of my sleep processes represents a process that might run too long. The other sleep process is the timer. Either process could play either role, but I envisioned that first line as the timer and the second as the process to be timed. One of the processes will probably finish first and then run the "kill -term -$$". This will kill the process group. On a multiprocessor system simultaneous kills may occur, but the kernel will ensure that they happen serially. There is no race condition here. This depends on the shell putting all processes in a script into a single process group. So neither sleep process could be replaced with another shell script because it would become a new process group.
 

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KILL(1) 						      General Commands Manual							   KILL(1)

NAME
kill, broke - print commands to kill processes SYNOPSIS
kill name broke DESCRIPTION
Kill prints commands that will cause all processes called name and owned by the current user to be terminated. Use the send command of 81/2(1), or pipe the output of kill into rc(1) to execute the commands. Kill suggests sending a kill note to the process; the same message delivered to the process's ctl file (see proc(3)) is a surer, if heavy handed, kill, but is necessary if the offending process is ignoring notes. Broke prints commands that will cause all processes in the Broken state and owned by the current user to go away. When a process dies because of an error caught by the system, it may linger in the Broken state to allow examination with a debugger. Executing the commands printed by broke lets the system reclaim the resources used by the broken processes. SOURCE
/rc/bin/kill /rc/bin/broke SEE ALSO
ps(1), stop(1), proc(3) KILL(1)
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