See some of the previous discussions on asyncio reentrancy.
https://mail.python.org/pipermail/async-sig/2019-March/thread.html
I do think there is some case for non rentrency and nested loop where
what you define here would block an outer loop, but most people
suggesting what you ask actually want re-entrency, which is not
possible there.
In your case test() would fail in a context where there is already a
running loo.
Also Explicit is better than implicit.
--
Matthias
On Fri, 12 Jun 2020 at 13:50, J. Pic
Hi all,
Currently, you can not use await outside an async function, the following code:
async def lol(): return 'bar'
def test(): return await lol()
print(test())
Will fail with: SyntaxError: 'await' outside async function
Of course, you can use asyncio.run and then it works fine:
import asyncio
async def lol(): return 'bar'
def test(): return asyncio.run(lol())
print(test())
Why not make using await do asyncio.run "behind the scenes" when called outside async function ?
Thank you in advance for your replies
-- ∞ _______________________________________________ Python-ideas mailing list -- python-ideas@python.org To unsubscribe send an email to python-ideas-leave@python.org https://mail.python.org/mailman3/lists/python-ideas.python.org/ Message archived at https://mail.python.org/archives/list/python-ideas@python.org/message/LOCYSY... Code of Conduct: http://python.org/psf/codeofconduct/