How do I get results from the blocking function? - python-asyncio

Question

import asyncio
import time
def abab():
for i in range(10):
print(i)
time.sleep(1)
return 10000
async def ab():
while True:
print(111)
time.sleep(1)
await asyncio.sleep(0.01)
async def main():
abc = asyncio.create_task(ab())
loop = asyncio.get_event_loop()
a = loop.run_in_executor(None, abab)
await abc
asyncio.run(main())
while print(111) of ab() function
After 10 seconds, I want to get abab result 10000.
abab function must be blocking function. not a async function
this example print like this
0 111 1 111 2 111 3 111 4 111 5 111 6 111 7 111 8 111 9 111 111 111 111...
but i want to print 10000 when return result
0 111 1 111 2 111 3 111 4 111 5 111 6 111 7 111 8 111 9 111 10000 111 111 111...
This is actually an example, but In the actual code, abab is a blocking function and cannot be modified. I can't think of a solution to this. I need help.

Answer 1

If you await the result of run_in_executor from somewhere, you will receive the return value of the blocking function it executed. Since you already use create_task to run ab() in the background, nothing stops you from just awaiting the call to run_in_executor to pick up the result of abab():
async def main():
# spawn ab() in the background
abc = asyncio.create_task(ab())
loop = asyncio.get_event_loop()
# wait for abab() to finish, allowing ab() to run
abab_result = await loop.run_in_executor(None, abab)
# print (or otherwise process) the result of abab
print(abab_result)
# continue waiting for ab()
await abc
With this change, the output from the program matches the one you requested.