Bash – sleep, wait and Ctrl+C propagation

bashsleepwait

If I have the following shell script

sleep 30s

And I hit Ctrl+C when the shell script is running, the sleep dies with it.

If I have the following shell script

sleep 30s &
wait

And I hit Ctrl+C when the shell script is running, the sleep continues on, and now has a parent of 1.

Why is that? Doesn't bash propagate Ctrl+C to all the children?

EDIT:
If I have the following script

/usr/bin/Xvfb :18.0 -ac -screen 0 1180x980x24 &
wait

where I am spawning a program, this time Ctrl+C on the main process kills the Xvfb process too.

So how/why is Xvfb different from sleep?

In the case of some processes I see that they get reaped by init, in some cases they die. Why does sleep get reaped by init? Why does Xvfb die?

Best Answer

tl;dr; the Xvfb process sets a signal handler for SIGINT and exits when it receives such a signal, but the sleep process doesn't, so it inherits the "ignore" state for SIGINT as it was set by the shell running the script before executing the sleep binary.

When a shell script is run, the job control is turned off, and background processes (the ones started with &) are simply run in the same process group, with SIGINT and SIGQUIT set to SIG_IGN (ignored) and with their stdin redirected from /dev/null.

This is required by the standard:

If job control is disabled (see the description of set -m) when the shell executes an asynchronous list, the commands in the list shall inherit from the shell a signal action of ignored (SIG_IGN) for the SIGINT and SIGQUIT signals.

If the signal disposition is set to SIG_IGN (ignore), that state will be inherited through fork() and execve():

Signals set to the default action (SIG_DFL) in the calling process image shall be set to the default action in the new process image. Except for SIGCHLD, signals set to be ignored (SIG_IGN) by the calling process image shall be set to be ignored by the new process image.

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