I wrote a simple bash script which does nothing but sleeps.
#!/bin/bash
echo "Sleeping..."
sleep 180s
I see two processes running on my system after I run the script:
user 22880 0.0 0.0 12428 1412 pts/28 S+ 20:12 0:00 /bin/bash ./sleep.sh
user 22881 0.0 0.0 7196 356 pts/28 S+ 20:12 0:00 sleep 180s
I give a SIGTERM to the process with id 22880 by using kill -15 22880 which kills the process. However, after this, I still see the sleep command running which exits after 180 seconds.
user 22881 0.0 0.0 7196 356 pts/28 S 20:12 0:00 sleep 180s
Why does this happen? What do I need to do to not leave the sleep 180s process running?
kill -15 22880 will send a signal to the shell executing the script, but not the sleep command. To send the signal to every process in the process group, you should be able to specify a negative process ID.
kill -15 -22880
Alternately, ensure that the script kills its children before exiting.
trap 'kill $(jobs -p)' EXIT
echo "Sleeping..."
sleep 180s & wait
If you leave sleep in the foreground when the signal is received, the shell must wait until it exits before running the trap; sleep is uninterruptible. The workaround is to run it in the background, then wait on it. wait, being a shell built-in, can be interrupted, so that the trap runs immediately and kills any background processes still in progress.