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

solution to wait for several subprocesses and to exit when any one of them exits with non-zero status code is by using 'wait -n'

#!/bin/bash
wait_for_pids()
{
    for (( i = 1; i <= $#; i++ )) do
        wait -n $@
        status=$?
        echo "received status: "$status
        if [ $status -ne 0 ] && [ $status -ne 127 ]; then
            exit 1
        fi
    done
}

sleep_for_10()
{
    sleep 10
    exit 10
}

sleep_for_20()
{
    sleep 20
}

sleep_for_10 &
pid1=$!

sleep_for_20 &
pid2=$!

wait_for_pids $pid2 $pid1

status code '127' is for non-existing process which means the child might have exited.