我试图在在线REPL here中重现该错误。但是,它与我的真实代码(在REPL中执行async for response in position_stream()
而不是for position in count()
)中的实现和行为并不完全相同。
我在某处定义一个协程,如下所示:
async def position(self):
request = telemetry_pb2.SubscribePositionRequest()
position_stream = self._stub.SubscribePosition(request)
try:
async for response in position_stream:
yield Position.translate_from_rpc(response)
finally:
position_stream.cancel()
其中position_stream是无限的(或可能持续很长时间)。我从这样的示例代码中使用它:
async def print_altitude():
async for position in drone.telemetry.position():
print(f"Altitude: {position.relative_altitude_m}")
和print_altitude()
在循环中使用:
asyncio.ensure_future(print_altitude())
asyncio.get_event_loop().run_forever()
那很好。现在,在某个时候,我想关闭来自调用方的流。我以为我可以只运行asyncio.ensure_future(loop.shutdown_asyncgens())
并等待靠近我的finally
被叫,但那没有发生。
相反,我收到有关未检索到的异常的警告:
Task exception was never retrieved
future: <Task finished coro=<print_altitude() done, defined at [...]
为什么会这样,如何使所有异步生成器实际上都关闭(并运行其finally
子句)?
答案 0 :(得分:1)
首先,如果您stop
进行循环,则协程中没有一个将有机会正常关闭。调用close
基本上意味着不可逆转地破坏循环。
如果您不关心正在运行的任务会发生什么,只需将它们全部cancel
都将停止,这也会停止异步生成器:
import asyncio
from contextlib import suppress
async def position_stream():
while True:
await asyncio.sleep(1)
yield 0
async def print_position():
async for position in position_stream():
print(f'position: {position}')
async def cleanup_awaiter():
await asyncio.sleep(3)
print('cleanup!')
if __name__ == '__main__':
loop = asyncio.get_event_loop()
try:
asyncio.ensure_future(print_position())
asyncio.ensure_future(print_position())
loop.run_until_complete(cleanup_awaiter())
# get all running tasks:
tasks = asyncio.gather(*asyncio.Task.all_tasks())
# schedule throwing CancelledError into the them:
tasks.cancel()
# allow them to process the exception and be cancelled:
with suppress(asyncio.CancelledError):
loop.run_until_complete(tasks)
finally:
print('closing loop')
loop.close()