proper way to detect shell exit code when errexit option set

Question

I prefer to write solid shell code, so the errexit & nounset is alway set.

The following code will stop at bad_command line

#!/bin/b         


        

Answer 1

In case you want to detect the exit code of a compound list (or a function), and have errexit and nounset applied there, you can use this kind of code:

#!/bin/sh
set -eu
unset -v unbound_variable

f() {
    echo "before false"
    false
    echo "after false"
}

g() {
    echo "before unbound"
    var=$unbound_variable
    echo "after unbound"
}

set +e
(set -e; f)
echo error code of f = $?
set -e

echo still in main program

set +e
(set -e; g)
echo error code of g = $?
set -e

echo still in main program

The above should print non-zero error codes for both functions f and g. I suppose this works by any POSIX shell. You can also detect the error code in EXIT trap, but the shell exits thereafter. The problem with other proposed methods is that the errexit setting is ignored when an exit status of such a compound list is tested. Here is a quote from the POSIX standard:

The -e setting shall be ignored when executing the compound list following the while, until, if, or elif reserved word, a pipeline beginning with the ! reserved word, or any command of an AND-OR list other than the last.

Note that if you have defined your function like

f() (
    set -e
    ...
)

it is enough to do

set +e
f
echo exit code of f = $?
set -e

to get the exit code.