Proper way to retrieve return value of wait command

Question

I have the following code in a bash script:

 for job in `jobs -p`; do
     echo "PID => ${job}"
     if ! wait ${job} ; then
     echo "At least one test failed with exit code => $?" ;
     EXIT_CODE=1;
     fi
 done

wait <pid>, in this case, wait ${job}, should return the exit code, but I don't know how to get/see the value. What I need to do is change the above script to something like this:

for job in `jobs -p`; do
     echo "PID => ${job}"
     CODE=0;
     wait ${job} || CODE=$?
     if ! ${CODE} ; then
     echo "At least one test failed with exit code => ${CODE}" ;
     EXIT_CODE=1;
     fi
 done

is there a cleaner less verbose way to do this? I am confused about how to retrieve the "return value" from wait <pid>, besides using the technique I just used. Is there another way to do it?

codeforester · Answer 1 · 2017-02-14T19:10:01.323

2

How about this:

declare -A failures
for job in `jobs -p`; do
  wait "$job" || ((failures[$job] = $?))
done

if (( "${#failures[@]}" > 0 )); then
  # traverse the hash and print which processes failed, along with exit code
fi

edited Feb 14 '17 at 19:10

answered Feb 14 '17 at 05:44

codeforester

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Why do you need `{}` around `printf`? – Inian Feb 14 '17 at 05:48
@GrishaLevit - thanks for pointing it out. Updated the answer. – codeforester Feb 14 '17 at 19:10

Grisha Levit · Answer 2 · 2017-02-14T06:01:28.397

2

If you really want to be more concise, you could do like:

wait $job || {
   echo "$job failed with code $?"
   EXIT_CODE=1
}

Side Note: for job in $(jobs -p); do might not be the best idea. The jobs list is expanded when the loop starts, but jobs might complete while the loop is running (e.g. while you are waiting for the first job in the list, the second job finishes).

Instead, you could do something like:

while :; do
   wait -n && continue            # wait for any job to finish
   code=$?
   ((code==127)) && break         # no jobs found to wait for
   echo "A test failed with code $code"
done

edited Feb 14 '17 at 06:01

answered Feb 14 '17 at 05:48

Grisha Levit

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Congrats. Your side note is even more important then the original problem. – linuxfan says Reinstate Monica Feb 14 '17 at 07:14
If multiple jobs complete between calls to `wait -n`, you might miss a job that failed. – chepner Feb 14 '17 at 11:51

score 1 · Answer 3 · answered Feb 14 '17 at 05:55

1

Cleaner? What could be cleaner than the most obvious and straightforward way:

wait $job
status=$?
if (( $status != 0 )); then
    ...
fi

Sure it's a tiny bit verbose, but it's also easy to understand. What you don't want is code that will make you scratch your head and be confused.

answered Feb 14 '17 at 05:55

Joel Cornett

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yeah I didn't know you could do that, with status=$? on the following line after the wait call, capturing the result of the wait call – Alexander Mills Feb 14 '17 at 06:11

Proper way to retrieve return value of wait command

3 Answers3