I have a script that needs to launch a program in order for the program to create and populate some folders. The program only needs to run briefly and then close, before the script continues.
However when the program is killed, the script does not continue. How can I do this?
printf "before"
(timeout -sHUP 3s firefox)
printf "after"
I thought I was launching the program in a subshell but i am not sure if this is the case.
I solved this with a hacky work around.
set -o errexit
printf "before"
timeout -sHUP 3s firefox || true
printf "after"
The alternative cleaner method would be
set -o errexit
printf "before"
set +e
timeout -sHUP 3s firefox
set -e
printf "after"
|| : is a common idiom to suppress the exit code of a failing command when you don't care.printf takes 2 arguments - a FORMAT string, then an ARGUMENT (i.e the text to be outputted).
When you call printf "before" you are actually passing a format string with no argument, which is not displayed properly.
The easiest fix is to instead use echo:
echo "before"
timeout -sHUP 3s firefox
echo "after"
Or if you need formatting, to instead do e.g.:
printf "%s\n" "before"
timeout -sHUP 3s firefox
printf "%s\n" "after"
printf will work fine with one argument - he just didn't include a newline, which unlike echo it doesn't add by default. printf "before\n"
timeout. The issue seems to be in the use of set -o errexit
timeoutreturns an non-zero exit status when it kills the program that it runs. Ifset -e(or, equivalently,set -o errexit) is in effect when that happens, the script will stop immediately and silently. The parentheses ((...)) ensure that the program is run in a subshell. That doesn't change the effect of theerrexitsetting, and it's not clear why a subshell is being used here.set -o errexitset indeed. Any way to ignore the timeout error specifically but keeperrexit?