为什么这个异常会立即从 asyncio 任务中引发？

Question

我对文档的理解是 asyncio.Tasks 作为 asyncio.Future 子类，将存储其中引发的异常，我可以在闲暇时检索它们。

但是，在此示例代码中，立即引发了异常：

import asyncio

async def bad_task():
    raise Exception()

async def test():
    loop = asyncio.get_event_loop()
    task = loop.create_task(bad_task())
    await task

    # I would expect to get here
    exp = task.exception()
    # but we never do because the function exits on line 3

loop = asyncio.get_event_loop()
loop.run_until_complete(test())
loop.close()

示例输出 (Python 3.6.5):

python3 ./test.py
Traceback (most recent call last):
  File "./test.py", line 15, in <module>
    loop.run_until_complete(test())
  File "/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/asyncio/base_events.py", line 468, in run_until_complete
    return future.result()
  File "./test.py", line 9, in test
    await task
  File "./test.py", line 4, in bad_task
    raise Exception()
Exception

这是在异步代码中创建和调用任务的怪癖吗？

Answer 1

await 将引发任务抛出的任何异常，因为它旨在使异步代码看起来几乎与同步代码完全一样。如果你想抓住它们，你可以使用普通的 try...except 子句。

Answer 2

正如 Matti 所解释的，协程引发的异常会传播到等待站点。这是有意为之的，因为它确保错误不会在默认情况下静默传递。但是，如果需要这样做，肯定可以等待任务完成而不立即访问其 result/exception。

这是一个简单而有效的方法，通过使用一个小的中间体 Future:

async def test():
    loop = asyncio.get_event_loop()
    task = loop.create_task(bad_task())
    task_done = loop.create_future()  # you could also use asyncio.Event
    # Arrange for task_done to complete once task completes.
    task.add_done_callback(task_done.set_result)

    # Wait for the task to complete. Since we're not obtaining its
    # result, this won't raise no matter what bad_task() does...
    await task_done
    # ...and this will work as expected.
    exp = task.exception()

为什么这个异常会立即从 asyncio 任务中引发？

Why is this exception immediately raised from an asyncio Task?

python-3.x

python-asyncio