Bash linux: start multiple program in parallel and stop all when one has finished

Question

I'm working on Ubuntu and I want to create a bash file that do these things:
Launch one program (prog0) on core 1.
Wait 3 seconds.
Then log the information of CPU and Memory usage for prog0 (I use two instances of pidstat on core 0 for logging that information).
Then start another program (prog1) on core 0.
When prog1 has finished (I think that prog1 exits automatically), I would like to exit from all the previous process (prog0 and two pidstat).

taskset -c 1 prog0 -option0 &
sleep 3
taskset -c 0 pidstat 1 -C prog0 -u > log2 &
taskset -c 0 pidstat 1 -C prog0 -r > log3 &
taskset -c 0 prog1 -option1 > log1 

I don't know how to exit or kill all the processes started when prog1 has finished.

Answer 1

Add:

trap 'kill $(jobs -p)' EXIT

to the beginning of your script. This will kill all background jobs when your script exits.

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To create a script open a new file and paste the following into it:

#!/bin/bash
trap 'kill $(jobs -p)' EXIT
taskset -c 1 prog0 -option0 &
sleep 3
taskset -c 0 pidstat 1 -C prog0 -u > log2 &
taskset -c 0 pidstat 1 -C prog0 -r > log3 &
taskset -c 0 prog1 -option1 > log1

Save the file as runme.sh.

Make it executable: chmod +x runme.sh

Run it by executing: ./runme.sh or to run it in the background: ./runme.sh &

Now, when the last command taskset -c 0 prog1 -option1 > log1 has finished, the script will exit and it will kill all the background processes that it started.