asyncio: Is it possible to cancel a future been ru

问题:

I would like to start a blocking function in an Executor using the asyncio call loop.run_in_executor and then cancel it later, but that doesn't seem to be working for me.

Here is the code:

import asyncio
import time

from concurrent.futures import ThreadPoolExecutor


def blocking_func(seconds_to_block):
    for i in range(seconds_to_block):
        print('blocking {}/{}'.format(i, seconds_to_block))
        time.sleep(1)

    print('done blocking {}'.format(seconds_to_block))


@asyncio.coroutine
def non_blocking_func(seconds):
    for i in range(seconds):
        print('yielding {}/{}'.format(i, seconds))
        yield from asyncio.sleep(1)

    print('done non blocking {}'.format(seconds))


@asyncio.coroutine
def main():
    non_blocking_futures = [non_blocking_func(x) for x in range(1, 4)]
    blocking_future = loop.run_in_executor(None, blocking_func, 5)
    print('wait a few seconds!')
    yield from asyncio.sleep(1.5)

    blocking_future.cancel()
    yield from asyncio.wait(non_blocking_futures)



loop = asyncio.get_event_loop()
executor = ThreadPoolExecutor(max_workers=1)
loop.set_default_executor(executor)
asyncio.async(main())
loop.run_forever()

I would expect the code above to only allow the blocking function to output:

blocking 0/5
blocking 1/5

and then see the output of the non blocking function. But instead the blocking future continues on even after I have canceled.

Is it possible? Is there some other way of doing it?

Thanks

Edit: More discussion on running blocking and non-blocking code using asyncio: How to interface blocking and non-blocking code with asyncio

回答1:

In this case, there is no way to cancel the Future once it has actually started running, because you're relying on the behavior of concurrent.futures.Future, and its docs state the following:

cancel()

Attempt to cancel the call. If the call is currently being executed and cannot be cancelled then the method will return False, otherwise the call will be cancelled and the method will return True.

So, the only time the cancellation would be successful is if the task is still pending inside of the Executor. Now, you're actually using an asyncio.Future wrapped around a concurrent.futures.Future, and in practice the asyncio.Future returned by loop.run_in_executor() will raise a CancellationError if you try to yield from it after you call cancel(), even if the underlying task is actually already running. But, it won't actually cancel the execution of the task inside the Executor.

If you need to actually cancel the task, you'll need to use a more conventional method of interrupting the task running in the thread. The specifics of how you do that is use-case dependent. For the use-case you presented in the example, you could use a threading.Event:

def blocking_func(seconds_to_block, event):
    for i in range(seconds_to_block):
        if event.is_set():
            return
        print('blocking {}/{}'.format(i, seconds_to_block))
        time.sleep(1)

    print('done blocking {}'.format(seconds_to_block))


...
event = threading.Event()
blocking_future = loop.run_in_executor(None, blocking_func, 5, event)
print('wait a few seconds!')
yield from asyncio.sleep(1.5)

blocking_future.cancel()  # Mark Future as cancelled
event.set() # Actually interrupt blocking_func

回答2:

As threads share the same memory address space of a process, there is no safe way to terminate a running thread. This is the reason why most programming languages do not allow to kill running threads (there are lots of ugly hacks around this limitation).

Java learnt it the hard way.

A solution would consist in running your function in a separate process instead of a thread and terinate it gracefully.

The Pebble library offers an interface similar to concurrent.futures supporting running Futures to be cancelled.

from pebble import ProcessPool

def function(foo, bar=0):
    return foo + bar

with ProcessPool() as pool:
    future = pool.schedule(function, args=[1])

    # if running, the container process will be terminated 
    # a new process will be started consuming the next task
    future.cancel()