[Python] pycharm 被装饰过的函数参数提示问题

B = TypeVar("B", bound="Bot")
R = TypeVar("R")
P = ParamSpec("P")

class API(Generic[B, P, R]):
    def __init__(self, func: Callable[Concatenate[B, P], Awaitable[R]]) -> None:
        self.func = func

    def __set_name__(self, owner: Type[B], name: str) -> None:
        self.name = name

    @overload
    def __get__(self, obj: None, objtype: Type[B]) -> "API[B, P, R]": ...

    @overload
    def __get__(
        self, obj: B, objtype: Optional[Type[B]]
    ) -> Callable[P, Awaitable[R]]: ...

    def __get__(
        self, obj: Optional[B], objtype: Optional[Type[B]] = None
    ) -> "API[B, P, R] | Callable[P, Awaitable[R]]":
        if obj is None:
            return self

        return partial(obj.call_api, self.name)  # type: ignore

    async def __call__(self, inst: B, *args: P.args, **kwds: P.kwargs) -> R:
        return await self.func(inst, *args, **kwds)
    

可以参考这个库的代码 https://github.com/nonebot/adapter-dodo/blob/83d55b09a9787c55d06fb01a5bb9c1f2d6a61fde/nonebot/adapters/dodo/utils.py#L39

像下面这个函数 https://github.com/nonebot/adapter-dodo/blob/83d55b09a9787c55d06fb01a5bb9c1f2d6a61fde/nonebot/adapters/dodo/bot.py#L298

被上面这个装饰器装饰过后就没有参数提示了... 在不换 IDE 的情况下该怎么做😭😭😭