我有如下代码:
from typing import Callable
def decorate(func: Callable[[str], None]) -> Callable[[str], None]:
return func
@decorate
def do_something(some_str: str = 'Hello world') -> None:
print(some_str)
if __name__ == '__main__':
do_something()
运行mypy时，它会在最后一行报告以下错误:
错误:"do\u something"[调用参数]的参数太少
如何修复此错误(不将返回类型decorate更改为Callable[..., None])？
使用TypeVar表示接收到的any Callable的输入和返回类型等效.由于Callable个类型包含整个签名，因此类型判断器也可以推断位置/可选/关键字参数.
from typing import Callable, TypeVar
# A type variable that can represent any Callable type
C = TypeVar("C", bound=Callable)
# The input Callable type defines the output Callable type as well
def decorate(func: C) -> C:
return func
如果decorate只能处理可能需要字符串的可调用项，则相应地调整bound.
# A type variable that can represent any Callable type that takes a string
C = TypeVar("C", bound=Callable[[str], None])
值得注意的是，接受str or using a default的可调用函数仍然满足Callable[[str], None]，因为它可以用str参数调用.