reliable return code of background process

Question

Let's assume the following piece of bash code:

foo > logfile 2>&1 &
foo_pid=$!

while ps -p$foo_pid
do
    ping -c 1 localhost
done

wait $foo_pid

if [[ $? == 0 ]]
then
    echo "foo success"
fi

Is it safe to assume that $? indeed contains the return code of foo and not the return code of ping? If the answer to that question is: "You cannot assume that." then how can I modify this piece of code to be sure that $? always contains the return code of foo?

Solution

With bash, you'll have that guarantee unless you've started another background job (and beware that background jobs can be started with & but also with coproc and with process substitution) between the foo & and the wait.

POSIX requires that a shell remembers the exit status of at least 25 jobs after they're gone, but bash remembers a lot more than that.

Now, if you do:

foo & pid=$!
...
bar &
wait "$pid"

You've got no guarantee that bar will not be given the same pid as foo (if foo has terminated by the time bar starts), so even though it's unlikely, wait "$pid" may give you the exit status of bar.

You can reproduce it with:

bash -c '(exit 12; foo) & pid=$!
         while : bar & [ "$pid" != "$!" ]; do :;done
         wait "$pid"; echo "$?"'

which will (eventually) give you 0 instead of 12.

To avoid the problem, one way would be to write it as:

{
  foo_pid=$!

  while ps -p "$foo_pid"
  do
      ping -c 1 localhost
  done

  bar &
  ...

  read <&3 ret
  if [ "$ret" = 0 ]; then
    echo foo was sucessful.
  fi
} 3< <(foo > logfile 2>&1; echo "$?")

Comments

[Stéphane Chazelas]

No, see my comment to umlaute's answer, and try it for yourself with bash -c '(exit 12) & sleep 1; wait "$!"; echo "$?"'

[Stéphane Chazelas]

I've never seen bash loose track (even after starting thousands of jobs), my example was demonstrating the pid being reused, which may be what you observed in your case as well.