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 ?

S

Answer 1

I've had a go at this and combined all the best parts from the other examples here. This script will execute the checkpids function when any background process exits, and output the exit status without resorting to polling.

#!/bin/bash

set -o monitor

sleep 2 &
sleep 4 && exit 1 &
sleep 6 &

pids=`jobs -p`

checkpids() {
    for pid in $pids; do
        if kill -0 $pid 2>/dev/null; then
            echo $pid is still alive.
        elif wait $pid; then
            echo $pid exited with zero exit status.
        else
            echo $pid exited with non-zero exit status.
        fi
    done
    echo
}

trap checkpids CHLD

wait

Answer 2

There are already a lot of answers here, but I am surprised no one seems to have suggested using arrays... So here's what I did - this might be useful to some in the future.

n=10 # run 10 jobs
c=0
PIDS=()

while true

    my_function_or_command &
    PID=$!
    echo "Launched job as PID=$PID"
    PIDS+=($PID)

    (( c+=1 ))

    # required to prevent any exit due to error
    # caused by additional commands run which you
    # may add when modifying this example
    true

do

    if (( c < n ))
    then
        continue
    else
        break
    fi
done 


# collect launched jobs

for pid in "${PIDS[@]}"
do
    wait $pid || echo "failed job PID=$pid"
done

Answer 3

Just store the results out of the shell, e.g. in a file.

#!/bin/bash
tmp=/tmp/results

: > $tmp  #clean the file

for i in `seq 0 9`; do
  (doCalculations $i; echo $i:$?>>$tmp)&
done      #iterate

wait      #wait until all ready

sort $tmp | grep -v ':0'  #... handle as required

Answer 4

This is something that I use:

#wait for jobs
for job in `jobs -p`; do wait ${job}; done

Answer 5

How about simply:

#!/bin/bash

pids=""

for i in `seq 0 9`; do
   doCalculations $i &
   pids="$pids $!"
done

wait $pids

...code continued here ...

Update:

As pointed by multiple commenters, the above waits for all processes to be completed before continuing, but does not exit and fail if one of them fails, it can be made to do with the following modification suggested by @Bryan, @SamBrightman, and others:

#!/bin/bash

pids=""
RESULT=0


for i in `seq 0 9`; do
   doCalculations $i &
   pids="$pids $!"
done

for pid in $pids; do
    wait $pid || let "RESULT=1"
done

if [ "$RESULT" == "1" ];
    then
       exit 1
fi

...code continued here ...

Answer 6

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.