pythonpython-3.xasync-awaitpython-asynciocoroutine

python asyncio add_done_callback with async def


I have 2 functions: The first one, def_a, is an asynchronous function and the second one is def_b which is a regular function and called with the result of def_a as a callback with the add_done_callback function.

My code looks like this:

import asyncio

def def_b(result):
    next_number = result.result()
    # some work on the next_number
    print(next_number + 1)

async def def_a(number):
    await some_async_work(number)
    return number + 1

loop = asyncio.get_event_loop()
task = asyncio.ensure_future(def_a(1))
task.add_done_callback(def_b)
response = loop.run_until_complete(task)
loop.close()

And it's work perfectly.

The problem began when also the second function, def_b, became asynchronous. Now it looks like this:

async def def_b(result):
    next_number = result.result()
    # some asynchronous work on the next_number
    print(next_number + 1)

But now I can not provide it to the add_done_callback function, because it's not a regular function.

My question is- Is it possible and how can I provide def_b to the add_done_callback function if def_b is asynchronous?


Solution

  • add_done_callback is considered a "low level" interface. When working with coroutines, you can chain them in many ways, for example:

    import asyncio
    
    
    async def my_callback(result):
        print("my_callback got:", result)
        return "My return value is ignored"
    
    
    async def coro(number):
        await asyncio.sleep(number)
        return number + 1
    
    
    async def add_success_callback(fut, callback):
        result = await fut
        await callback(result)
        return result
    
    
    loop = asyncio.get_event_loop()
    task = asyncio.ensure_future(coro(1))
    task = add_success_callback(task, my_callback)
    response = loop.run_until_complete(task)
    print("response:", response)
    loop.close()
    

    Keep in mind add_done_callback will still call the callback if your future raises an exception (but calling result.result() will raise it).