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 just been modifying a script to background and parallelise a process.

I did some experimenting (on Solaris with both bash and ksh) and discovered that 'wait' outputs the exit status if it's not zero , or a list of jobs that return non-zero exit when no PID argument is provided. E.g.

Bash:

$ sleep 20 && exit 1 &
$ sleep 10 && exit 2 &
$ wait
[1]-  Exit 2                  sleep 20 && exit 2
[2]+  Exit 1                  sleep 10 && exit 1

Ksh:

$ sleep 20 && exit 1 &
$ sleep 10 && exit 2 &
$ wait
[1]+  Done(2)                  sleep 20 && exit 2
[2]+  Done(1)                  sleep 10 && exit 1

This output is written to stderr, so a simple solution to the OPs example could be:

#!/bin/bash

trap "rm -f /tmp/x.$$" EXIT

for i in `seq 0 9`; do
  doCalculations $i &
done

wait 2> /tmp/x.$$
if [ `wc -l /tmp/x.$$` -gt 0 ] ; then
  exit 1
fi

While this:

wait 2> >(wc -l)

will also return a count but without the tmp file. This might also be used this way, for example:

wait 2> >(if [ `wc -l` -gt 0 ] ; then echo "ERROR"; fi)

But this isn't very much more useful than the tmp file IMO. I couldn't find a useful way to avoid the tmp file whilst also avoiding running the "wait" in a subshell, which wont work at all.