Catch errors in asyncio.ensure_future

问题:

I have this code:

try:
    asyncio.ensure_future(data_streamer.sendByLatest())
except ValueError as e:
    logging.debug(repr(e))

data_streamer.sendByLatest() can raise a ValueError, but it is not caught.

回答1:

ensure_future - just creates Task and return immediately. You should await for created task to get it's result (including case when it raises exception):

import asyncio


async def test():
    await asyncio.sleep(0)
    raise ValueError('123')


async def main():    
    try:
        task = asyncio.ensure_future(test())  # Task aren't finished here yet 
        await task  # Here we await for task finished and here exception would be raised 
    except ValueError as e:
        print(repr(e))


if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    loop.run_until_complete(main())

Output:

ValueError('123',)

In case you aren't planning to await task immediately after you created it, you can await it later (to know how it has finished):

async def main():    
    task = asyncio.ensure_future(test())
    await asyncio.sleep(1)
    # At this moment task finished with exception,
    # but we didn't retrieved it's exception.
    # We can do it just awaiting task:
    try:
        await task  
    except ValueError as e:
        print(repr(e)) 

Output is same:

ValueError('123',)