python - asyncio 在任务中捕获 TimeoutError

Question

我有一个 asyncio.Task 需要在一段时间后取消。在取消之前，任务需要做一些清理。根据文档，我应该只能调用 task.cancel 或 asyncio.wait_for(coroutine, delay) 并拦截协程中的 asyncio.TimeoutError，但是以下示例不起作用。我试过拦截其他错误，并改为调用 task.cancel，但都没有用。我是否误解了取消任务的工作原理？

@asyncio.coroutine
def toTimeout():
  try:
    i = 0
    while True:
      print("iteration ", i, "......"); i += 1
      yield from asyncio.sleep(1)
  except asyncio.TimeoutError:
    print("timed out")

def main():
  #... do some stuff
  yield from asyncio.wait_for(toTimeout(), 10)
  #... do some more stuff

asyncio.get_event_loop().run_until_complete(main())
asyncio.get_event_loop().run_forever()

Answer 1

documentation for asyncio.wait_for指定它将取消底层任务，然后从 wait_for 调用自身引发 TimeoutError:

Returns result of the Future or coroutine. When a timeout occurs, it cancels the task and raises asyncio.TimeoutError.

你是正确的任务取消 can indeed be intercepted :

[Task.cancel] arranges for a CancelledError to be thrown into the wrapped coroutine on the next cycle through the event loop. The coroutine then has a chance to clean up or even deny the request using try/except/finally.

请注意，文档指定 CancelledError 被抛入协程，而不是 TimeoutError。

如果您进行了调整，事情就会按照您预期的方式进行:

import asyncio

@asyncio.coroutine
def toTimeout():
  try:
    i = 0
    while True:
      print("iteration ", i, "......"); i += 1
      yield from asyncio.sleep(1)
  except asyncio.CancelledError:
    print("timed out")

def main():
  #... do some stuff
  yield from asyncio.wait_for(toTimeout(), 3)
  #... do some more stuff

asyncio.get_event_loop().run_until_complete(main())

输出:

iteration  0 ......
iteration  1 ......
iteration  2 ......
timed out
Traceback (most recent call last):
  File "aio.py", line 18, in <module>
    asyncio.get_event_loop().run_until_complete(main())
  File "/usr/lib/python3.4/asyncio/base_events.py", line 316, in run_until_complete
    return future.result()
  File "/usr/lib/python3.4/asyncio/futures.py", line 275, in result
    raise self._exception
  File "/usr/lib/python3.4/asyncio/tasks.py", line 238, in _step
    result = next(coro)
  File "aio.py", line 15, in main
    yield from asyncio.wait_for(toTimeout(), 3)
  File "/usr/lib/python3.4/asyncio/tasks.py", line 381, in wait_for
    raise futures.TimeoutError()
concurrent.futures._base.TimeoutError

如您所见，现在 'timed out' 在 wait_for 引发 TimeoutError 之前打印。