Reputation: 5
queue = []
async def funcA():
r = random.randint(0, 1000)
i = 0
for i in range(10000000):
pass
print(r)
queue.append(r)
asyncio.create_task(funcb(r))
async def funcb(r):
i = 0
for i in range(100000000):
pass
print(r, "XX")
queue.pop()
async def main():
for i in range(10):
await funcA()
print(queue)
if __name__ == "__main__":
asyncio.run(main())
How do I make funcb()
run simultaneously? In this code funcb()
runs only after all calls of funcA()
are done executing. I want funcB()
to run concurrently with funcA()
(on same or different threads). Here if funcA()
runs infinitely, then funcB()
would never run.
Upvotes: 0
Views: 683
Reputation: 22456
Asyncio is a so-called non-preemptive task scheduler.
This means, it cannot switch back and forth between tasks on its own, it requires the function it is calling to actively give up (=yield) the priority.
As your function is not asynchronous (= doesn't have any commands in it that would wait for something with the await
keyword), it just runs to completion before yielding back to the asyncio scheduler.
The way you described your problem sounds like it isn't really suitable for asynchronous programming, but instead requires multithreading or even multiprocessing (if you desire to get a speedup from the 'simultaneous' computing).
Upvotes: 4