Why Bash is like that: Signal propagation

Bash can seem pretty random and weird at times, but most of what people see as quirks have very logical (if not very good) explanations behind them. This series of posts looks at some of them.

How do I simulate pressing Ctrl-C when running this in a script:
while true; do echo sleeping; sleep 30; done

Are you thinking “SIGINT, duh!”? Hold your horses!

I tried kill -INT pid, but it doesn't work the same:

Ctrl-C    kills the sleep and the loop
SIGINTing the shell does nothing (but only in scripts: see Errata)
SIGINTing sleep makes the loop continue with the next iteration

HOWEVER, if I run the script in the background and kill -INT %1
instead of kill -INT pid, THEN it works :O

Why does Ctrl-C terminate the loop, while SIGINT doesn’t?

Additionally, if I run the same loop with ping or top instead of sleep,
Ctrl-C doesn't terminate the loop either!

Yeah. Well… Yeah…

This behaviour is due to an often overlooked feature in UNIX: process groups. These are important for getting terminals and shells to work the way they do.

A process group is exactly what it sounds like: a group of processes. They have a leader, which is the process that created it using setsid(2). The leader’s pid is also the process group id. Child processes are in the same group as their parent by default.

Terminals keep track of the foreground process group (set by the shell using tcsetpgrp(3)). When receiving a Ctrl-C, they send the SIGINT to the entire foreground group. This means that all members of the group will receive SIGINT, not just the immediate process.

kill -INT %1 sends the signal to the job’s process group, not the backgrounded pid! This explains why it works like Ctrl-C.

You can do the same thing with kill -INT -pgrpid. Since the process group id is the same as the process group leader, you can kill the group by killing the pid with a minus in front.

But why do you have to kill both?

When the shell is interrupted, it will wait for the running command to exit. If this child’s status indicates it exited abnormally due to that signal, the shell cleans up, removes its signal handler, and kills itself again to trigger the OS default action (abnormal exit). Alternatively, it runs the script’s signal handler as set with trap, and continues.

If the shell is interrupted and the child’s status says it exited normally, then Bash assumes the child handled the signal and did something useful, so it continues executing. Ping and top both trap SIGINT and exit normally, which is why Ctrl-C doesn’t kill the loop when calling them.

This also explains why interrupting just the shell does nothing: the child exits normally, so the shell thinks the child handled the signal, though in reality it was never received.

Finally, if the shell isn’t interrupted and a child exits, Bash just carries on regardless of whether the signal died abnormally or not. This is why interrupting the sleep just continues with the loop.

In case one would like to handle such cases, Bash sets the exit code to 128+signal when the process exits abnormally, so interrupting sleep with SIGINT would give the exit code 130 (kill -l lists the signal values).

Bonus problem:

I have this C app, testpg:
int main() {
    return sleep(10);

I run bash -c './testpg' and press Ctrl-C. The app is killed.
Shouldn't testpg be excluded from SIGINT, since it used setsid?

A quick strace unravels this mystery: with a single command to execute, bash execve’s it directly — a little optimization trick. Since the pid is the same and already had its own process group, creating a new one doesn’t have any effect.

This trick can’t be used if there are more commands, so bash -c './testpg; true' can’t be killed with Ctrl-C.


Wait, I started a loop in one terminal and killed the shell in another. 
The loop exited!

Yes it did! This does not apply to interactive shells, which have different ways of handling signals. When job control is enabled (running interactively, or when running a script with bash -m), the shell will die when SIGINTed

Here’s the description from the bash source code, jobs.c:2429:

  /* Ignore interrupts while waiting for a job run without job control
     to finish.  We don't want the shell to exit if an interrupt is
     received, only if one of the jobs run is killed via SIGINT. 

4 thoughts on “Why Bash is like that: Signal propagation”

  1. \SIGINTing sleep makes the loop continue with the next iteration\

    Actually, for me it stops the iteration.

  2. In a Debian 7/32bit x86 machine, and from a bash 4.3.25 shell – which happens to have a PID of 3701 – I run this:

    (typing at the command line, not via a script)
    “while true ; do echo Sleeping ; sleep 1 ; done”.

    If from another terminal I send an INT signal to the shell (“kill -INT 3701”) the loop is indeed, aborted – so this is not what you described above (“SIGINTing the shell does nothing”).

    I guess bash has changed in the interim?

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