Bash: wait with timeout

Question

In a Bash script, I would like to do something like:

app1 &
pidApp1=$!
app2 &
pidApp2=$1

timeout 60 wait $pidApp1 $pidApp2
kill -9 $pidApp1 $pidApp2         


        

Answer 1

Both your example and the accepted answer are overly complicated, why do you not only use timeout since that is exactly its use case? The timeout command even has an inbuilt option (-k) to send SIGKILL after sending the initial signal to terminate the command (SIGTERM by default) if the command is still running after sending the initial signal (see man timeout).

If the script doesn't necessarily require to wait and resume control flow after waiting it's simply a matter of

timeout -k 60s 60s app1 &
timeout -k 60s 60s app2 &
# [...]

If it does, however, that's just as easy by saving the timeout PIDs instead:

pids=()
timeout -k 60s 60s app1 &
pids+=($!)
timeout -k 60s 60s app2 &
pids+=($!)
wait "${pids[@]}"
# [...]

E.g.

$ cat t.sh
#!/bin/bash

echo "$(date +%H:%M:%S): start"
pids=()
timeout 10 bash -c 'sleep 5; echo "$(date +%H:%M:%S): job 1 terminated successfully"' &
pids+=($!)
timeout 2 bash -c 'sleep 5; echo "$(date +%H:%M:%S): job 2 terminated successfully"' &
pids+=($!)
wait "${pids[@]}"
echo "$(date +%H:%M:%S): done waiting. both jobs terminated on their own or via timeout; resuming script"

.

$ ./t.sh
08:59:42: start
08:59:47: job 1 terminated successfully
08:59:47: done waiting. both jobs terminated on their own or via timeout; resuming script