Sync tasks in loop

In this example too we’ll use sleep to pretend we are waiting for some external task to finish, but this time we’ll start a number of jobs based on what the user supplies. We can’t know up-front how many tasks we’ll have to call.

Code

import time
import sys

def do_task(task_id: int, sec: int):
    print(f"Start task {task_id}")
    time.sleep(sec)
    print(f"End task {task_id}")

def main():
    if len(sys.argv) != 2:
        exit(f"Usage {sys.argv[0]} NUMBER")

    for i in range(int(sys.argv[1])):
        do_task(i, 1)


start = time.monotonic()
main()
end = time.monotonic()
print(f"Elapsed time: {end-start}")

Output

Start task 0
End task 0
Start task 1
End task 1
Start task 2
End task 2
Start task 3
End task 3
Elapsed time: 4.000992348068394

As one could expect from such code, the total time required for such program to run is the sum of all the tasks as they run sequentially.