How can I get exit status of subshell?

Question

A classical example of piping:

$ false | true
$ echo ${PIPESTATUS[@]}
1 0

Same example run in a subshell and assigning results into a variable:

$ process="$(false | true)"
$ echo ${PIPESTATUS[@]}
0

How could I get the exit status of each of the piped processes?

Similar question, without the subshell part:

`process="$(false | true ; echo ${PIPESTATUS[@]} >&2)"` or `process="$(false | true ; echo ${PIPESTATUS[@]} >/dev/tty)"` — jesse_b, Nov 16 '19 at 16:55
well, you do the `echo ${PIPESTATUS}` _inside_ the command substitution subshell ;-) I guess your question is more about how to capture output other than stdout from a command substitution. — , Nov 16 '19 at 16:56
... which is "interesting", since there are no nice solutions, only ugly kludges. For instance, this will use a tempfile to get the data out, while leaving stdout and stderr undisturbed: `t=$(mktemp); exec 3>$t 4<$t; rm "$t"; out=$(false | true | false | echo hmm; echo -n "${PIPESTATUS[@]}" >&3); readarray -td' ' st <&4; exec 3>&- 4<&-; echo "out=$out; st=(${st[*]})"`. Related: [Redirect STDERR and STDOUT to different variables without temporary files](https://unix.stackexchange.com/questions/430161/redirect-stderr-and-stdout-to-different-variables-without-temporary-files) — , Nov 16 '19 at 17:22
btw, you got you quoting backwards: you should quote the `${PIPESTATUS[@]}` in `echo "${PIPESTATUS[@]}"`, but you don't have to quote the `$(...)` in `process=$(false | true)`. — , Nov 16 '19 at 17:26

0 Answers0