Propagating exit code to caller in case of a shell error from script having an exit trap

Question

Is it possible to propagate an exit code to the caller in case of a syntax error in a Bash script with an EXIT trap? For example, if I have:

#! /bin/bash

set -eu

trap "echo dying!!" EXIT

echo yeah
echo $UNBOUND_VARIABLE
echo boo

Then, running it gives an exit code 0 even if the script did not really end successfully:

$ bash test.sh
yeah
test.sh: line 8: UNBOUND_VARIABLE: unbound variable
dying!!

$ echo $?
0

But if I comment out the exit trap, the script returns 1. Alternatively, if I replace the line with the unbound variable with a command that returns nonzero (e.g. /bin/false), that exit value is propagated as I would like it to.

Answer 1

The shell exits with the result of the last executed command. In your trap case, that's echo, which usually returns with success.

To propagate your value, simply exit with it.

#!/bin/bash

set -eu

die() {
  echo "Dying!!"
  exit "$1"
}

trap 'die $?' EXIT

echo yeah
echo $unbound
echo boo

Also note that set -e is considered harmful -- it makes you think the script will exit if a command fails, which it won't always do.

Answer 2

This behavior is related to different Bash versions. The original script works as expected on Bash 4.2 but not on 3.2. Having the error-prone code in a separate script file and running it in a subshell works around problems in earlier Bash versions:

#!/bin/bash

$BASH sub.sh
RETVAL=$?

if [[ "$RETVAL" != "0" ]]; then
  echo "Dying!! Exit code: $RETVAL"
fi

sub.sh:

set -eu

echo yeah
echo $UNBOUND_VARIABLE
echo boo