asyncio 在任务中捕获 TimeoutError

Eva*_*mas 3 python python-3.x python-asyncio

我有asyncio.Task一段时间需要取消。在取消之前,任务需要做一些清理工作。根据文档,我应该只能在协程中调用 task.cancel 或asyncio.wait_for(coroutine, delay)拦截 an asyncio.TimeoutError,但以下示例不起作用。我尝试过拦截其他错误并调用task.cancel,但都没有成功。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()
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dan*_*ano 6

文档asyncio.wait_for指定它将取消基础任务,然后TimeoutErrorwait_for调用本身引发:

返回 Future 或协程的结果。当发生超时时,它会取消任务并引发asyncio.TimeoutError

你是对的,任务取消确实可以被拦截

[ Task.cancel] 安排CancelledError在事件循环的下一个循环中将 a 扔到包装的协程中。try然后,协程有机会使用//except清理甚至拒绝请求 finally

请注意,文档指定CancelledError将 放入协程中,而不是TimeoutError.

如果您进行调整,事情就会按照您期望的方式进行:

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())
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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
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正如您所看到的, now在引发'timed out'之前被打印。TimeoutErrorwait_for