装饰函数返回类型在直接调用时保持原始

Question

定义一个接受Callable[..., int]并返回Callable[..., str]的装饰器似乎无法被mypy理解

def decorator(wrapped: Callable[..., int]) -> Callable[..., str]:
    def wrapper() -> str:
        return str(wrapped())

    return wrapper


@decorator
def foo() -> int:
    return 0


def bar():
    x = foo()  # mypy sees x as int even though the decorator(foo)() returns str

有什么办法可以让foo()的返回类型不是它原来的返回类型，而是装饰器指定的返回类型？

编辑：简化示例

Answer 1

这是我在最新版本的 mypy (0.930) 中看到的：str not int

def decorator(wrapped: Callable[..., int]) -> Callable[..., str]:

    def wrapper() -> str:
        return str(wrapped())

    return wrapper


@decorator
def foo() -> int:
    return 0


def bar() -> None:
    x = foo()
    reveal_type(x)  # Mypy: Revealed type is "builtins.str"