是什么导致asyncio.new_event_loop()的简单调用挂起?

时间:2016-03-08 07:11:24

标签: python hang coroutine python-asyncio event-loop

我正在使用以下函数强制协程同步运行:

import asyncio
import inspect
import types
from asyncio import BaseEventLoop
from concurrent import futures


def await_sync(coro: types.CoroutineType, timeout_s: int=None):
    """

    :param coro: a coroutine or lambda loop: coroutine(loop)
    :param timeout_s:
    :return:
    """
    loop = asyncio.new_event_loop()  # type: BaseEventLoop
    if not is_awaitable(coro):
        coro = coro(loop)
    if timeout_s is None:
        fut = asyncio.ensure_future(coro, loop=loop)
    else:
        fut = asyncio.ensure_future(asyncio.wait_for(coro, timeout=timeout_s, loop=loop), loop=loop)
    loop.run_until_complete(fut)
    return fut.result()

def is_awaitable(coro_or_future):
    if isinstance(coro_or_future, futures.Future):
        return coro_or_future
    elif asyncio.coroutines.iscoroutine(coro_or_future):
        return True
    elif asyncio.compat.PY35 and inspect.isawaitable(coro_or_future):
        return True
    else:
        return False

然而,间歇性地,只要尝试创建一个新循环它就会冻结:loop = asyncio.new_event_loop()。检查堆栈跟踪向我显示它挂起的确切位置:

File: "/src\system\utils.py", line 34, in await_sync
  loop = asyncio.new_event_loop()  # type: BaseEventLoop
File: "\lib\asyncio\events.py", line 636, in new_event_loop
  return get_event_loop_policy().new_event_loop()
File: "\lib\asyncio\events.py", line 587, in new_event_loop
  return self._loop_factory()
File: "\lib\asyncio\selector_events.py", line 55, in __init__
  self._make_self_pipe()
File: "\lib\asyncio\selector_events.py", line 116, in _make_self_pipe
  self._ssock, self._csock = self._socketpair()
File: "\lib\asyncio\windows_events.py", line 295, in _socketpair
  return windows_utils.socketpair()
File: "\lib\socket.py", line 515, in socketpair
  ssock, _ = lsock.accept()
File: "\lib\socket.py", line 195, in accept
  fd, addr = self._accept()

socket等低级别的库中可能导致此类问题的原因是什么?难道我做错了什么?我使用的是Python 3.5.1。

编辑:我提交了错误报告here,但Guido建议我继续寻求StackOverflow的帮助。

1 个答案:

答案 0 :(得分:1)

我正试图了解你要做什么。

如果我正确理解它,那么无论输入是协程还是简单的函数调用,您都希望有一个函数可以返回相同的结果。如果我是正确的,那么这似乎很好。

import asyncio
import time

def await_sync(coro, timeout=None):
  if asyncio.iscoroutine(coro):
    f = asyncio.wait_for(coro, timeout)
    loop = asyncio.get_event_loop()
    return loop.run_until_complete(f)
  return coro

async def async_test(x):
    print("x", end="")
    await asyncio.sleep(x)
    print("y", end="")
    return x

def sync_test(x):
    print("x", end="")
    time.sleep(x)
    print("y", end="")
    return x

print(await_sync(sync_test(2)))
print(await_sync(async_test(3)))

这将输出以下(我想是预期的)结果:

xy2
xy3