exit from bash script that uses $()

Question

I was under impression that exit would terminate the current bash script no matter what, and had the following error handler at the top of my script:

function err {
    printf "\e[31m$1\e[0m\n" 1>&2
    exit 1
}

It worked like a charm for most cases until this line:

item=$(myfunc $1)

Normally, that line works fine, with the STDOUT of myfunc dumped into $item, as intended. The problem arises when myfunc throws an error, via the err function seen above. The $() ends up swallowing the non-zero return and guarding the exit from exiting the script itself. If I understand correctly, the problem is that $() actually spawns a new subshell (just like the deprecated backticks), but I know of no other way to capture the output of a function into a variable in bash that allows the exit to work.

I tried using set -e as well, and had no luck with that either. Can someone suggest how to build my error handler so that it exits the script even in these cases?

Answer 1

You can test the result of the assignment:

if item=$(myfunc "$1")
then : Function worked
else : Function failed
fi

This tests the exit status of the command run in the sub-shell that the $(…) uses.

---

Without actually using functions, you can experiment with:

$ if item=$(echo Hi; exit 1); then echo "$item - OK"; else echo "$item - OH"; fi
Hi - OH
$ if item=$(echo Hi; exit 0); then echo "$item - OK"; else echo "$item - OH"; fi
Hi - OK
$

Or, if functions are deemed crucial, then:

$ err() { exit $1; }
$ myfunc() { echo Mine; err $1; }
$ if item=$(myfunc 1); then echo "$item - OK"; else echo "$item - OH"; fi
Mine - OH
$ if item=$(myfunc 0); then echo "$item - OK"; else echo "$item - OH"; fi
Mine - OK
$

Tested using Bash 3.2.57 on Mac OS X 10.11 El Capitan.