Execute several programs at the same time in an initialisation/bash script

Question

Hello I am working with a simulator that uses rcS scripts to boot, this is my script

cd /tests
./test1 &
./test2 &
./test3 &
./test4 
exit

What I want is run all the test at the same time and that the exit command is executed only when all the previous test have finished. And not only when test 4 has finished, is this possible?. Thank you.

Answer 1

You can use wait:

./test1 &
./test2 &
./test3 &
./test4 &
wait

From the bash man page:

wait [n ...] Wait for each specified process and return its termination status. Each n may be a process ID or a job specification; if a job spec is given, all processes in that job's pipeline are waited for. If n is not given, all currently active child processes are waited for, and the return status is zero. If n specifies a non-existent process or job, the return status is 127. Otherwise, the return status is the exit status of the last process or job waited for.

Answer 2

xargs can support parallel

So just like this:

seq 4|xargs -i -n 1 -P 4 ./test{} 

Answer 3

Something along the lines of

cd /tests
./test1 &
./test2 &
./test3 &
./test4 &
wait
exit

(I am assuming bash shell)