Python type hints and context managers

How should a context manager be annotated with Python type hints?

import typing

@contextlib.contextmanager
def foo() -> ???:
    yield

The documentation on contextlib doesn't mention types much.

The documentation on typing.ContextManager is not all that helpful either.

There's also typing.Generator, which at least has an example. Does that mean I should use typing.Generator[None, None, None] and not typing.ContextManager?

import typing

@contextlib.contextmanager
def foo() -> typing.Generator[None, None, None]:
    yield

Solution

Whenever I'm not 100% sure what types a function accepts, I like to consult typeshed, which is the canonical repository of type hints for Python. Mypy directly bundles and uses typeshed to help it perform its typechecking, for example.

We can find the stubs for contextlib here: https://github.com/python/typeshed/blob/master/stdlib/contextlib.pyi

if sys.version_info >= (3, 2):
    class GeneratorContextManager(ContextManager[_T], Generic[_T]):
        def __call__(self, func: Callable[..., _T]) -> Callable[..., _T]: ...
    def contextmanager(func: Callable[..., Iterator[_T]]) -> Callable[..., GeneratorContextManager[_T]]: ...
else:
    def contextmanager(func: Callable[..., Iterator[_T]]) -> Callable[..., ContextManager[_T]]: ...

It's a little overwhelming, but the line we care about is this one:

def contextmanager(func: Callable[..., Iterator[_T]]) -> Callable[..., ContextManager[_T]]: ...

It states that the decorator takes in a Callable[..., Iterator[_T]] -- a function with arbitrary arguments returning some iterator. So in conclusion, it would be fine to do:

@contextlib.contextmanager
def foo() -> Iterator[None]:
    yield

So, why does using Generator[None, None, None] also work, as suggested by the comments?

It's because Generator is a subtype of Iterator -- we can again check this for ourselves by consulting typeshed. So, if our function returns a generator, it's still compatible with what contextmanager expects so mypy accepts it without an issue.