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

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

# run processes and store pids in array
for i in $n_procs; do
    ./procs[${i}] &
    pids[${i}]=$!
done

# wait for all pids
for pid in ${pids[*]}; do
    wait $pid
done

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