In bash, when running ( sleep 123 &), the sleep 123 process will continue running, when the subshell exits. How can I stop the sleep 123 process before its parent subshell exits?

I'm trying to see if the sleep 123 process will be terminated, because of receiving SIGHUP and SIGCONT. I am looking for an example for Is SIGHUP sent to this orphaned process, and why doesn't it terminate? and Does kernel sending SIGHUP to a process group that becomes orphaned and contains a stopped process terminate all the processes by default?

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    Is sleep just an example, or is this your real code? Because sleep (the binary) basically calls sleep (the kernel function) which means it only processes signals once time is up. – nohillside Dec 22 '18 at 16:45
  • Thanks. Do you mean sleep is not stoppable by signal? you can use whatever command that can be stopped. – Tim Dec 22 '18 at 17:01
  • Processes can't be stopped while they run in kernel mode. sleep 123 basically makes a syscall to sleep(123) which only will return once time has run out. So a real life example would call either another shell script or a non-sleeping binary inside the (...&). – nohillside Dec 22 '18 at 17:08
  • Thanks. If a "non-sleeping binary" makes system calls and a signal tries to stop the process while the system call is running, will it be stopped only after the system call returns to user space code? – Tim Dec 22 '18 at 17:14
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    @nohillside no, sleep is interruptible by non-ignored signals. – Stephen Kitt Dec 22 '18 at 17:44

This will show the behaviour you’re trying to illustrate:

(sleep 60 & kill -STOP $!)

This puts sleep in the background, then stops it. It then gets killed by SIGHUP when the subshell exits.

Signals can interrupt some system calls; see the signal(7) manpage (in particular the “Interruption of system calls and library functions by signal handlers” section). The system calls used by sleep in particular are interrupted when a signal handler is invoked, and this is documented in sleep(3).

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