使用Ctrl-C终止程序而不终止父脚本

Question

I have a bash script that starts an external program ( evtest ) twice.

#!/bin/bash

echo "Test buttons on keyboard 1"
evtest /dev/input/event1

echo "Test buttons on keyboard 2"
evtest /dev/input/event2

As far as I know, evtest can be terminated only via Ctrl-C. The problem is that this terminates the parent script, too. That way, the second call to evtest will never happen.

How can I close the first evtest without closing the script, so that the second evtest will actually run?

Thanks!

PS: for the one that want to ask "why not running evtest manually instead of using a script?", the answer is that this script contains further semi-automated hardware debug test, so it is more convenient to launch the script and do everything without the need to run further commands.

Answer 1

You can use the trap command to "trap" signals; this is the shell equivalent of the signal() or sigaction() call in C and most other programming languages to catch signals.

The trap is reset for subshells, so the evtest will still act on the SIGINT signal sent by ^C (usually by quiting), but the parent process (ie the shell script) won't.

Simple example:

#!/bin/sh

# Run a command command on signal 2 (SIGINT, which is what ^C sends)
sigint() {
    echo "Killed subshell!"
}
trap sigint 2

# Or use the no-op command for no output
#trap : 2

echo "Test buttons on keyboard 1"
sleep 500

echo "Test buttons on keyboard 2"
sleep 500

And a variant which still allows you to quit the main program by pressing ^C twice in a second:

last=0
allow_quit() {
    [ $(date +%s) -lt $(( $last + 1 )) ] && exit
    last=$(date +%s)
}
trap allow_quit 2