How to trap exit code in Bash script

The accepted answer is basically correct, I just want to clarify things.

The following example works well:

#!/bin/bash

cleanup() {
    rv=$?
    rm -rf "$tmpdir"
    exit $rv
}

tmpdir="$(mktemp)"
trap "cleanup" EXIT
# Do things...

But you have to be more careful if doing cleanup inline, without a function. For example this won't work:

trap "rv=$?; rm -rf $tmpdir; exit $rv" EXIT

Instead you have to escape the $rv and $? variables:

trap "rv=\$?; rm -rf $tmpdir; exit \$rv" EXIT

You might also want to escape $tmpdir, as it will get evaluated when the trap line gets executed and if the tmpdir value changes later that might not give the expected behaviour.

Edit: Use shellcheck to check your bash scripts and be aware of problems like this.

I think you can use $? to get the exit code.

I've found it is better to separate EXIT trap from the trap for other signals

Example trap test script...

umask 77
tmpfile=`tmpfile.$$`
trap 'rm -f "$tmpfile"' EXIT
trap 'exit 2' HUP INT QUIT TERM

touch $tmpfile
read -r input 

exit 10

The temporary file is cleaned up. The file exit value of 10 is preserved! Interrupts result in an exit value of 2

Basically as long as you don't use "exit" in a EXIT trap, it will exit with the original exit value preserved.

ASIDE: Note the quoting in the EXIT trap. That lets me change what file needs to be cleaned up during the scripts lifetime. I often also include a test for the existence of the $tmpfile before trying to remove it, so I don't even need to set it at the start of the script, only before creating it.