Issue 33117: asyncio example uses non-existing/documented method (original) (raw)

Issue33117

Created on 2018-03-21 21:06 by hfingler, last changed 2022-04-11 14:58 by admin. This issue is now closed.

Messages (3)
msg314223 - (view)	Author: Henrique Fingler (hfingler)	Date: 2018-03-21 21:06
In the documentation of asyncio.run_coroutine_threadsafe(coro, loop), in Section 19.5.3.6 (https://docs.python.org/3/library/asyncio-task.html#asyncio.run_coroutine_threadsafe), the example code does the following: future = asyncio.run_coroutine_threadsafe(coro, loop) # Wait for the result with an optional timeout argument assert future.result(timeout) == 3 The problem is that the result method of a future, according to the documentation doesn't take parameters. It's in Section 19.5.3.4 (https://docs.python.org/3.8/library/asyncio-task.html#asyncio.Future.done) result() Return the result this future represents. The same function is used in Section 18.5.9.3 (https://docs.python.org/3/library/asyncio-dev.html#concurrency-and-multithreading) This error is present in all Python 3.* docs. From the asyncio source code (https://github.com/python/cpython/blob/master/Lib/asyncio/futures.py), we have this in the Future class definition: class Future: """This class is *almost* compatible with concurrent.futures.Future. Differences: - This class is not thread-safe. - result() and exception() do not take a timeout argument and raise an exception when the future isn't done yet. .... So this example needs to be reworked, I'd do it if I knew more about asyncio. My ideas involve either using a add_done_callback with a flag or just busy waiting until future.done().
msg325867 - (view)	Author: Karthikeyan Singaravelan (xtreak) *	Date: 2018-09-20 12:57
I think `asyncio.run_coroutine_threadsafe(coro, loop)` returns a concurrent.futures.Future [1] which supports timeout and not asyncio.Future which doesn't support timeout. Please add in if I am missing anything. Since asyncio docs are being rewritten this would be a great time to contribute too if you would like to add more clarification. Docs at [1] also say the below : asyncio.run_coroutine_threadsafe(coro, loop) Submit a coroutine to the given event loop. Thread-safe. Return a concurrent.futures.Future to wait for the result from another OS thread. ``` # bpo33117.py. This needs to be called from a different thread as docs said but I am using future.result(timeout) just to make sure there is no error with respect to function argument. import asyncio loop = asyncio.get_event_loop() timeout = 4 # Create a coroutine coro = asyncio.sleep(1, result=3) # Submit the coroutine to a given loop future = asyncio.run_coroutine_threadsafe(coro, loop) # Wait for the result with an optional timeout argument assert future.result(timeout) == 3 ``` ➜ cpython git:(master) ./python.exe bpo33117.py Traceback (most recent call last): File "../bpo33117.py", line 13, in assert future.result(timeout) == 3 File "/Users/karthikeyansingaravelan/stuff/python/cpython/Lib/concurrent/futures/_base.py", line 438, in result raise TimeoutError() concurrent.futures._base.TimeoutError [1] https://docs.python.org/3.8/library/concurrent.futures.html#concurrent.futures.Future Thanks
msg326795 - (view)	Author: Andrew Svetlov (asvetlov) *	Date: 2018-10-01 11:39
@xtreak is right, `run_coroutine_threadsafe()` returns `concurrent.futures.Future` object. Personally, I like the fact that `fut.result()` is called with timeout parameter to reflect the fact of the different object type.