CODEWITHSUNDEEP
pythonpython-3.xpython-asyncio
dpetican
dpetican

Reputation: 831

Python using futures with loop_forever

Just started experimenting with asynch which looks really cool. I'm trying to use futures with an asynch coroutine that runs forever but I get this error:

Task exception was never retrieved
future: <Task finished coro=<slow_operation() done, defined at ./asynchio-test3.py:5> exception=InvalidStateError("FINISHED: <Future finished result='This is the future!'>",)>

This is my code which runs as expected if I remove the 3 lines related to futures:

import asyncio

@asyncio.coroutine
def slow_operation():
    yield from asyncio.sleep(1)
    print ("This is the task!")
    future.set_result('This is the future!')
    asyncio.async(slow_operation())

def got_result(future):
    print(future.result())

loop = asyncio.get_event_loop()
future = asyncio.Future()
future.add_done_callback(got_result)
asyncio.async(slow_operation())
try:
    loop.run_forever()
finally:
    loop.close()

Upvotes: 2

Views: 1649

Answers (2)

dpetican
dpetican

Reputation: 831

Following @falsetru answer this is a complete program that has 3 asynch coroutines each with their own got_result function. I'm using v3.4 so thats why I don't use the new syntax. As an interesting side effect the output clearly demonstrates the single threaded nature of coroutines. I hope its useful as a template for someone:

import asyncio

@asyncio.coroutine
def task1(future):
    yield from asyncio.sleep(1)
    print ("This is operation#1")
    future.set_result('This is the result of operation #1!')
    asyncio.async(task1(new_future(got_result1)))

def got_result1(future):
    print(future.result())

@asyncio.coroutine
def task2(future):
    yield from asyncio.sleep(1)
    print ("This is operation#2")
    future.set_result('This is the result of operation #2!')
    asyncio.async(task2(new_future(got_result2)))

def got_result2(future):
    print(future.result())

@asyncio.coroutine
def task3(future):
    yield from asyncio.sleep(1)
    print ("This is operation#3")
    future.set_result('This is the result of operation #3!')
    asyncio.async(task3(new_future(got_result3)))

def got_result3(future):
    print(future.result())

def new_future(callback):
    future = asyncio.Future()
    future.add_done_callback(callback)
    return future

tasks = [task1(new_future(got_result1)),
        task2(new_future(got_result2)),
        task3(new_future(got_result3))]

loop = asyncio.get_event_loop()
for task in tasks:
    asyncio.async(task)

try:
    loop.run_forever()
finally:
    loop.close()

Upvotes: 1

falsetru
falsetru

Reputation: 368894

slow_operator is called indefinitely, calling set_result for the same future object multiple times; which is not possbile.

>>> import asyncio
>>> future = asyncio.Future()
>>> future.set_result('result')
>>> future.set_result('result')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "C:\Python35\lib\asyncio\futures.py", line 329, in set_result
    raise InvalidStateError('{}: {!r}'.format(self._state, self))
asyncio.futures.InvalidStateError: FINISHED: <Future finished result='result'>

Create new future for each slow_operator call. For example:

@asyncio.coroutine
def slow_operation(future):
    yield from asyncio.sleep(1)
    print ("This is the task!")
    future.set_result('This is the future!')
    asyncio.async(slow_operation(new_future()))

def got_result(future):
    print(future.result())

def new_future():
    future = asyncio.Future()
    future.add_done_callback(got_result)
    return future

loop = asyncio.get_event_loop()
asyncio.async(slow_operation(new_future()))
try:
    loop.run_forever()
finally:
    loop.close()

BTW, you can use new syntax (async, await) if you're using Python 3.5+:

async def slow_operation(future):
    await asyncio.sleep(1)
    print ("This is the task!")
    future.set_result('This is the future!')
    asyncio.ensure_future(slow_operation(new_future()))

Upvotes: 2

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