使用 asyncio
时,我在捕获从信号处理程序回调抛出的自定义异常时遇到问题。
如果我从下面的 do_io()
中抛出 ShutdownApp
,我就能够在 run_app()
中正确捕获它。但是,当从 handle_sig()
引发异常时,我似乎无法捕获它。
Minimal, Reproducible Example使用 Python 3.8.5 进行测试:
import asyncio
from functools import partial
import os
import signal
from signal import Signals
class ShutdownApp(BaseException):
pass
os.environ["PYTHONASYNCIODEBUG"] = "1"
class App:
def __init__(self):
self.loop = asyncio.get_event_loop()
def _add_signal_handler(self, signal, handler):
self.loop.add_signal_handler(signal, handler, signal)
def setup_signals(self) -> None:
self._add_signal_handler(signal.SIGINT, self.handle_sig)
def handle_sig(self, signum):
print(f"\npid: {os.getpid()}, Received signal: {Signals(signum).name}, raising error for exit")
raise ShutdownApp("Exiting")
async def do_io(self):
print("io start. Press Ctrl+C now.")
await asyncio.sleep(5)
print("io end")
def run_app(self):
print("Starting Program")
try:
self.loop.run_until_complete(self.do_io())
except ShutdownApp as e:
print("ShutdownApp caught:", e)
# TODO: do other shutdown related items
except:
print("Other error")
finally:
self.loop.close()
if __name__ == "__main__":
my_app = App()
my_app.setup_signals()
my_app.run_app()
print("Finished")
在异步 Debug模式下按CTRL+C
(对于SIGINT
)后的输出:
(env_aiohttp) anav@anav-pc:~/Downloads/test$ python test_asyncio_signal.py
Starting Program
io start. Press Ctrl+C now.
^C
pid: 20359, Received signal: SIGINT, raising error for exit
Exception in callback App.handle_sig(<Signals.SIGINT: 2>)
handle: <Handle App.handle_sig(<Signals.SIGINT: 2>) created at /home/anav/miniconda3/envs/env_aiohttp/lib/python3.8/asyncio/unix_events.py:99>
source_traceback: Object created at (most recent call last):
File "test_asyncio_signal.py", line 50, in <module>
my_app.setup_signals()
File "test_asyncio_signal.py", line 25, in setup_signals
self._add_signal_handler(signal.SIGINT, self.handle_sig)
File "test_asyncio_signal.py", line 22, in _add_signal_handler
self.loop.add_signal_handler(signal, handler, signal)
File "/home/anav/miniconda3/envs/env_aiohttp/lib/python3.8/asyncio/unix_events.py", line 99, in add_signal_handler
handle = events.Handle(callback, args, self, None)
Traceback (most recent call last):
File "/home/anav/miniconda3/envs/env_aiohttp/lib/python3.8/asyncio/events.py", line 81, in _run
self._context.run(self._callback, *self._args)
File "test_asyncio_signal.py", line 31, in handle_sig
raise ShutdownApp("Exiting")
ShutdownApp: Exiting
io end
Finished
预期输出:
Starting Program
io start. Press Ctrl+C now.
^C
pid: 20359, Received signal: SIGINT, raising error for exit
ShutdownApp caught: Exiting
io end
Finished
是否可以从asyncio
中的信号处理程序引发自定义异常?如果是这样,我如何正确捕获/排除它?
最佳答案
handle_sig
是一个回调,因此它直接在事件循环之外运行,并且其异常仅通过全局 Hook 报告给用户。如果您希望在程序的其他地方捕获那里引发的异常,则需要使用 future 将异常从 handle_sig
传输到您希望它注意到的位置。
要在顶层捕获异常,您可能需要引入另一个方法,我们将其称为 async_main()
,它会等待任一 self. do_io()
或之前创建的 future 来完成:
def __init__(self):
self.loop = asyncio.get_event_loop()
self.done_future = self.loop.create_future()
async def async_main(self):
# wait for do_io or done_future, whatever happens first
io_task = asyncio.create_task(self.do_io())
await asyncio.wait([self.done_future, io_task],
return_when=asyncio.FIRST_COMPLETED)
if self.done_future.done():
io_task.cancel()
await self.done_future # propagate the exception, if raised
else:
self.done_future.cancel()
要从handle_sig
内部引发异常,您只需set the exception关于 future 的对象:
def handle_sig(self, signum):
print(f"\npid: {os.getpid()}, Received signal: {Signals(signum).name}, raising error for exit")
self.done_future.set_exception(ShutdownApp("Exiting"))
最后,修改 run_app
以将 self.async_main()
传递给 run_until_complete
,一切就完成了:
$ python3 x.py
Starting Program
io start. Press Ctrl+C now.
^C
pid: 2069230, Received signal: SIGINT, raising error for exit
ShutdownApp caught: Exiting
Finished
最后,请注意,可靠地捕获键盘中断是 notoriously tricky undertaking并且上述代码可能无法涵盖所有极端情况。
关于python - 如何从 asyncio 中的信号处理程序捕获自定义异常?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/64227873/