PEP 484 typing -- 如何注解 Callable withOut 返回类型？

Question

是否可以使用 NoReturn 类型注释 Callable？

我希望这样做的方式会产生错误：

from sys import exit
from typing import Callable, NoReturn

f: Callable[..., NoReturn] = exit

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/Versions/3.7/lib/python3.7/typing.py", line 755, in __getitem__
    return self.__getitem_inner__(params)
  File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/Versions/3.7/lib/python3.7/typing.py", line 251, in inner
    return func(*args, **kwds)
  File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/Versions/3.7/lib/python3.7/typing.py", line 774, in __getitem_inner__
    result = _type_check(result, msg)
  File "/usr/local/Cellar/python/3.7.0/Frameworks/Python.framework/Versions/3.7/lib/python3.7/typing.py", line 135, in _type_check
    raise TypeError(f"Plain {arg} is not valid as type argument")
TypeError: Plain typing.NoReturn is not valid as type argument

编辑：对于以后遇到此问题的任何人，此问题是 Python 3.7.0 中的错误，升级到 Python 3.7.2 可缓解此问题。

Answer 1

对我来说（使用 Python 3.7.2）它可以正常工作：

>>> from sys import exit
>>> from typing import Callable, NoReturn
>>> f: Callable[..., NoReturn] = exit
>>>

PEP 484 typing -- 如何注解 Callable withOut 返回类型？

PEP484 typing -- how to annotate Callable with NoReturn type?

python

python-3.x

python-typing