Pass return value from coroutine to callback function without using await

Question

import asyncio

async def f():
    # something, for example await asyncio.sleep(4)
    return "return value from f"

async def main():
    # schedule f, print result, but don't await
    for i in range(10):
        await asyncio.sleep(1)  # just an example for some other task
        print(i)

asyncio.run(main())

I could use print(await f()) but that blocks due to await. I want print to be called as a "callback function" after f() has returned while the rest of main() has already continued. So, assuming that # something in f took 4 seconds, the expected output would look like this:

0
1
2
3
return value from f
4
5
6
7
8
9

You probably have a few options. add_done_callback, gather, and as_completed come to mind. Without knowing more specifics, though, it's hard to say which will work best. — dirn
– dirn, Commented Jul 12, 2021 at 16:42

PIG208 · Accepted Answer · 2021-07-12 17:31:48Z

1

You can define a function that takes f and a callback function as parameters.

async def call(f, cb):
    cb(await f())

And you can schedule a task for it without using await:

asyncio.create_task(call(f, print))

answered Jul 12, 2021 at 17:31

PIG208

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Leo Wotzak · Accepted Answer · 2021-07-12 17:00:27Z

0

Does this work for you? I changed the f() coroutine a little bit but this does seem to work. Also, you need to await asyncio.sleep(). Let me know what you think or if you have any questions.

Here's the answer that I referenced: How I call an async function without await?

import asyncio

async def f():
    await asyncio.sleep(3)
    print("return value from f")

async def main():
    loop = asyncio.get_event_loop()
    loop.create_task(f())
    for i in range(10):
        await asyncio.sleep(1)  # just an example for some other task
        print(i)

asyncio.run(main())

answered Jul 12, 2021 at 17:00

Leo Wotzak

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