Executing one function multiple times simultaneously

• target=func1() calls func1 and passes its return value (in this case None) as the target argument for Process instead of passing the function func1 itself.

• Process(...) just creates a Process object. You never really spawn/execute any of the created Process objects. You need to add a call to .start. Depending on the OS (ie if os.fork is used to spawn the new process), this will also require us to add __main__ guard (which is a best-practice anyway regardless of the used OS).

• You are using p1 twice.

Try this:

from multiprocessing import Process
import time

def func1():
    time.sleep(1)
    print('Finished sleeping')

if __name__ == '__main__':
    t1_start = time.perf_counter()
    p1 = Process(target=func1)  # note no ()
    p2 = Process(target=func1)  # note no ()

    p1.start()
    p2.start()

    t1_stop = time.perf_counter()

    print("elapsed time: {} sec".format(round(t1_stop - t1_start), 1))

This gives the output

elapsed time: 0 sec
Finished sleepingFinished sleeping

which makes sense. If the function is executed 2 times in separated processes then the main process, the main process would not execute time.sleep(1) at all.

However, if we add .join() then the main process will be forced to wait for the child processes:

p1.start()
p2.start()

p1.join()
p2.join()

Now the output is the same as your required output:

Finished sleeping
Finished sleeping
elapsed time: 1 sec

EDIT: If you want an arbitrary number of processes with a loop:

...
times = 3
processes = []
for _ in range(times):
    p = Process(target=func1)
    p.start()
    processes.append(p)

for p in processes:
    p.join()
...