In the case below, the report command must always be executed but I need to get an exit status 1 if the test command fails:

echo $?

How can I do it in a single bash line without creating a shell script?


Save and reuse $?.

test; ret=$?; report; exit $ret

If you have multiple test commands and you want to run them all, but keep track of whether one has failed, you can use bash's ERR trap.

trap 'failures=$((failures+1))' ERR
if ((failures == 0)); then
  echo "Success"
  echo "$failures failures"
  exit 1

What is a shell script except a file containing shell commands? You could pretend it's not a shell script and put it on one line with something like:

(test; r=$?; report; [ "$r" -gt 0 ] && exit 1; exit 0)

This exits the subshell with a 1 if the first command returned anything other than 0; otherwise, it returns 0.

For examples -- I'm just echoing something in lieu of running a real program:

$ (false; r=$?; echo report; [ "$r" -gt 0 ] && exit 1; exit 0)
$ echo $?
$ (true; r=$?; echo report; [ "$r" -gt 0 ] && exit 1; exit 0)
$ echo $?
$ # example with an exit-status of 2:
$ (grep foo bar; r=$?; echo report; [ "$r" -gt 0 ] && exit 1; exit 0)
grep: bar: No such file or directory
$ echo $?

If you regularly set the errexit shell option, you'd want to add in an override so that the subshell doesn't exit prematurely:

(set +o errexit; false; r=$?; echo report; [ "$r" -gt 0 ] && exit 1; exit 0)

You said "exit status 1 if the test command fails". Assuming you intended to make that distinction (i.e., you want the exit value to be 1 even if the test command exits with, say, 2, or 3, or whatever), then this is a decent solution:

test && { report; true; } || { report; false; }

If you did not really mean "1", but whatever non-zero that test exited with would be OK as your overall exit code, then the current top answer (save and reuse $?) works fine.


A really bad idea, but this seems to work, for some inspiration:

(trap report exit; test)

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