asyncio loop.run_until_complete not sleeping with asyncio.sleep

Question

here is the code:

In [168]: async def myTask():
     ...:     await asyncio.sleep(2)
     ...:     print("processing task")

In [168]: async def generator():
     ...:     for i in range(5):
     ...:         asyncio.create_task(myTask())

In [168]: def abcd():
     ...:     t0=time.time()
     ...:     loop=asyncio.get_event_loop()
     ...:     loop.run_until_complete(generator())
     ...:     t1=time.time()
     ...:     print(t1-t0)

Ideally calling abc() should sleep for around 10 seconds(as I have not awaited the asyncio.create_task(myTask())), but the function is getting completed in less than 0.1 secs.

In [167]: abcd()
processing task
processing task
processing task
processing task
processing task
0.0004038810729980469

You haven't awaited myTask(). If you try replacing myTask() with asyncio.sleep(2) directly, you'll get exactly the same behavior. Hmmm... — Mateen Ulhaq
– Mateen Ulhaq, Commented Jan 7, 2020 at 0:02
@MateenUlhaq Yes, but in this case, no output should be printed. It's confusing. — Harsh Sharma
– Harsh Sharma, Commented Jan 7, 2020 at 0:07
Did you save the file beforehand? Try it again with the sample you posted. It seems to generate no additional output for me. — Mateen Ulhaq
– Mateen Ulhaq, Commented Jan 7, 2020 at 0:09

Sraw · Accepted Answer · 2020-01-07 01:52:44Z

2

If you try the following code, you will get output after around 2 seconds. Well, but we haven't awaited any tasks. The reason is that the Jupyter notebook itself is running in an event loop. So when you use the default event loop, it actually points to the Jupyter notebook's event loop.

import asyncio

async def myTask():
    await asyncio.sleep(2)
    print("processing task")

for i in range(5):
    asyncio.create_task(myTask())

answered Jan 7, 2020 at 1:52

Sraw

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