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This question is close to others - Can I get the exit code from a sub shell launched with $(command)?

However there are no solutions I've found that allows me to get an exit code from a sub shell when using local and eval as in this example...

test() {
> local WHY="$(eval "echo 'test'"; exit 3)"; echo $?
> }
test
0
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  • 2
    there was one or more Qs where this was debated at length, but of course there's a trivial solution for that: local why; why=$(false); echo $?; (btw, don't use uppercase names for local variables).
    – user313992
    Commented Dec 28, 2019 at 7:10
  • that did the trick, thanks :)
    – openCivilisation
    Commented Dec 28, 2019 at 9:39
  • BTW, arguably, one shouldn't use all-uppercase for any shell variables that aren't defined by the shell itself -- see the relevant POSIX spec, specifying that the shell and other POSIX-compliant tools may modify their behavior based only on all-caps environment variables, leaving variable names with at least one lowercase character safe for applications to use (relevant to shell variables that aren't explicitly exported since changing a shell variable overwrites any like-named environment variable).
    – Charles Duffy
    Commented Dec 28, 2019 at 22:13

1 Answer 1

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9

This is simple: Do not use a single command but split:

test() {
    local why
    why="$(eval "echo 'test'"; exit 3)"; echo $?
}
test
3

The problem was that local is a builtin command with an own exit code...If you avoid that command at the same time as the variable assignment, you get the exit code from the subshell.

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