How to wait in bash for several subprocesses to finish and return exit code !=0 when any subprocess ends with code !=0 ?

Question

How to wait in a bash script for several subprocesses spawned from that script to finish and return exit code !=0 when any of the subprocesses ends with code !=0 ?

Simple script:

#!/bin/bash
for i in `seq 0 9`; do
  doCalculations $i &
done
wait

The above script will wait for all 10 spawned subprocesses, but it will always give exit status 0 (see help wait). How can I modify this script so it will discover exit statuses of spawned subprocesses and return exit code 1 when any of subprocesses ends with code !=0?

Is there any better solution for that than collecting PIDs of the subprocesses, wait for them in order and sum exit statuses?

Answer 1

wait also (optionally) takes the PID of the process to wait for, and with $! you get the PID of the last command launched in background. Modify the loop to store the PID of each spawned sub-process into an array, and then loop again waiting on each PID.

Answer 2

I don't believe it's possible with Bash's builtin functionality.

You can get notification when a child exits:

#!/bin/sh
set -o monitor        # enable script job control
trap 'echo "child died"' CHLD

However there's no apparent way to get the child's exit status in the signal handler.

Getting that child status is usually the job of the wait family of functions in the lower level POSIX APIs. Unfortunately Bash's support for that is limited - you can wait for one specific child process (and get its exit status) or you can wait for all of them, and always get a 0 result.

What it appears impossible to do is the equivalent of waitpid(-1), which blocks until any child process returns.

Answer 3

http://jeremy.zawodny.com/blog/archives/010717.html :

#!/bin/bash

FAIL=0

echo "starting"

./sleeper 2 0 &
./sleeper 2 1 &
./sleeper 3 0 &
./sleeper 2 0 &

for job in `jobs -p`
do
echo $job
    wait $job || let "FAIL+=1"
done

echo $FAIL

if [ "$FAIL" == "0" ];
then
echo "YAY!"
else
echo "FAIL! ($FAIL)"
fi

Answer 4

Here's what I've come up with so far. I would like to see how to interrupt the sleep command if a child terminates, so that one would not have to tune WAITALL_DELAY to one's usage.

waitall() { # PID...
  ## Wait for children to exit and indicate whether all exited with 0 status.
  local errors=0
  while :; do
    debug "Processes remaining: $*"
    for pid in "$@"; do
      shift
      if kill -0 "$pid" 2>/dev/null; then
        debug "$pid is still alive."
        set -- "$@" "$pid"
      elif wait "$pid"; then
        debug "$pid exited with zero exit status."
      else
        debug "$pid exited with non-zero exit status."
        ((++errors))
      fi
    done
    (("$#" > 0)) || break
    # TODO: how to interrupt this sleep when a child terminates?
    sleep ${WAITALL_DELAY:-1}
   done
  ((errors == 0))
}

debug() { echo "DEBUG: $*" >&2; }

pids=""
for t in 3 5 4; do 
  sleep "$t" &
  pids="$pids $!"
done
waitall $pids