python - __aexit__ 键盘中断

Question

__aenter__ 和 __aexit__ 方法旨在确保清理，但是在抛出键盘中断后，不再调用 __aexit__ 方法。我可以想象这是因为“with”语句不是当前执行点的堆栈跟踪的一部分(随意猜测)。如何在asyncio中有效处理这个问题？

编辑:我创建了一些示例代码:

import asyncio

class manager:
    async def __aenter__(self):
        print("enter")


    async def __aexit__(self, exc_type, exc_val, exc_tb):
        print("exit")

async def func1():
    print("func1")
    async with manager():
        await asyncio.sleep(1000)

async def func2():
    print("func2")
    await asyncio.sleep(2)
    raise KeyboardInterrupt

loop = asyncio.get_event_loop()
loop.create_task(func1())
loop.create_task(func2())
loop.run_forever()

Answer 1

您所有的任务都还在，您只需要决定如何处理它们。如果您有行为良好的任务，那么它应该像取消它们然后让它们运行直到完成一样简单。

例如。

try:
    loop.run_forever()
except BaseException:
    all_tasks = asyncio.Task.all_tasks(loop=loop)
    for task in all_tasks:
        task.cancel()
    loop.run_until_complete(
        asyncio.gather(
            *all_tasks,
            loop=loop,
            return_exceptions=True # means all tasks get a chance to finish
        )
    )
    raise
finally:
    loop.close()

产生:

func1
enter
func2
exit
Traceback (most recent call last):
  File "C:\Users\User\Documents\python\asyncio_test.py", line 26, in <module>
    loop.run_forever()
  File "C:\Users\User\AppData\Local\Programs\Python\Python35-32\lib\asyncio\base_events.py", line 295, in run_forever
    self._run_once()
  File "C:\Users\User\AppData\Local\Programs\Python\Python35-32\lib\asyncio\base_events.py", line 1254, in _run_once
    handle._run()
  File "C:\Users\User\AppData\Local\Programs\Python\Python35-32\lib\asyncio\events.py", line 125, in _run
    self._callback(*self._args)
  File "C:\Users\User\AppData\Local\Programs\Python\Python35-32\lib\asyncio\tasks.py", line 301, in _wakeup
    self._step()
  File "C:\Users\User\AppData\Local\Programs\Python\Python35-32\lib\asyncio\tasks.py", line 239, in _step
    result = coro.send(None)
  File "C:\Users\User\Documents\python\asyncio_test.py", line 19, in func2
    raise KeyboardInterrupt
KeyboardInterrupt

如果您有一个行为不佳的任务，那么此实现可能无法停止。一个这样的实现看起来像:

async def bad():
    while True:
        try:
            print('bad loop')
            await asyncio.sleep(0.5, loop=loop)
        except BaseException as e:
            print(repr(e))