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.
Solution 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
OTHER TIPS
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.