Sunday, February 27, 2022

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

Issue

How to wait in a bash script for several subprocesses spawned from that script to finish, and then 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?


Solution

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


Answered By - Luca Tettamanti
Answer Checked By - Candace Johnson (WPSolving Volunteer)
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