How to type function returning a class decorator? (python3.11)

Question

I know I can type a class decorator as follows:

T = TypeVar('T', bound=Type[Any])
def decorator(cls: T) -> T:
    ...

Which can be used:

@decorator
class Foo:
   foo: int

And a reference to foo on instance of Foo will check properly in mypy.

But how do I type a function that returns a decorator, so that the decorated class types correctly? E.g. how can I fix the code below, if the arguments to decorator have types not associated with the type of the wrapped class?

T = TypeVar('T', bound=Type[Any])
Decorator: TypeAlias = Callable[[T], T]

def decorator(...) -> Decorator[Any]:

    def wrapper(cls: T) -> T:
       # some code modifying cls depending on args to "decorator"
       ...
       return cls
    
    return wrapper

# now instance don't type write
@decorator(...)
class Foo:
    foo: int
    

Answer 1

Once the wrapper types become complex enough, I find that using typing.Protocol can be pretty useful. The function returned can be represented as non-generic class which supplies a generic __call__ method.

import typing as t

T = t.TypeVar('T', bound=t.Type[t.Any])


class Decorator(t.Protocol):
    def __call__(self, cls: T) -> T:
        ...


def decorator() -> Decorator:
    def wrapper(cls: T) -> T:
        # some code modifying cls depending on args to "decorator"
        ...
        return cls

    return wrapper


@decorator()
class Foo:
    foo: int


t.reveal_type(Foo)
t.assert_type(Foo(), Foo)

(venv) > mypy --strict main.py
main.py:25: note: Revealed type is "def () -> main.Foo"
Success: no issues found in 1 source file