無法獲取堆棧使用跟蹤asyncio.ensure_future（）

Question

這是我的嘗試：

>>> import asyncio 
>>> @asyncio.coroutine 
... def f(): 
... print("RUN") 
... raise Exception("ciao") 
... 
>>> asyncio.ensure_future(f()) 
<Task pending coro=<coro() running at /usr/lib/python3.4/asyncio/coroutines.py:139>> 
>>> loop = asyncio.get_event_loop() 
>>> loop.run_forever() 
RUN 

和堆棧跟蹤不獲取。如果我使用asyncio.run_until_complete(f())運行協程，則沒有問題。

Answer 1

您必須等待某個協程的結果，並且會在該上下文中引發異常。所有的協程都需要「等待」; asyncio.run_until_complete()會爲你隱式做，但run_forever()不能，因爲它應該永遠運行。下面是一個例子，你可以看到異常（使用Python 3.5語法）：

>>> import asyncio 
>>> import traceback 
>>> async def f(): 
...  raise Exception("Viva la revolución!") 
... 
>>> task_f = asyncio.ensure_future(f()) 
>>> async def g(): 
...  try: 
...   await task_f 
...  except Exception: 
...   traceback.print_exc() 
...   
>>> task_g = asyncio.ensure_future(g()) 
>>> asyncio.get_event_loop().run_forever() 
Traceback (most recent call last): 
    File "<ipython-input-5-0d9e3c563e35>", line 3, in g 
    await task_f 
    File "/usr/lib/python3.5/asyncio/futures.py", line 363, in __iter__ 
    return self.result() # May raise too. 
    File "/usr/lib/python3.5/asyncio/futures.py", line 274, in result 
    raise self._exception 
    File "/usr/lib/python3.5/asyncio/tasks.py", line 239, in _step 
    result = coro.send(None) 
    File "<ipython-input-3-928dc548dc3e>", line 2, in f 
    raise Exception("Viva la revolución!") 
Exception: Viva la revolución!