Callable 是无效的基类?

问题描述

谁能解释一下为什么要从未参数化和参数化的Callable中继承:

Can someone explain why inheriting from unparameterized and parameterized Callable:

from typing import Callable
from typing import NoReturn
from typing import TypeVar


T = TypeVar('T', str, int)
C = Callable[[T], NoReturn]


class Foo(Callable):

    def __call__(self, t: T):
        pass


class Bar(C):

    def __call__(self, t: T):
        pass

当传递给 mypy 时会引发 Foo 和 Bar 的错误:

when passed to mypy raises errors for both Foo and Bar:

tmp.py:13: error: Invalid base class
tmp.py:19: error: Invalid base class

推荐答案

这部分是因为运行时的类不能真正从函数或可调用的开始继承，部分是因为您不需要显式继承 Callable 以指示一个类是可调用的.

This is in part because classes at runtime can't really inherit from a function or a callable to begin with, and in part because you don't need to explicitly inherit from Callable to indicate that a class is callable.

例如，以下程序使用 mypy 0.630 按预期进行类型检查:

For example, the following program typechecks as expected using mypy 0.630:

from typing import Callable, Union, NoReturn, List

class Foo:
    def __call__(self, t: Union[str, int]) -> NoReturn:
        pass


class FooChild(Foo): pass


class Bad:
    def __call__(self, t: List[str]) -> NoReturn:
        pass


def expects_callable(x: Callable[[Union[str, int]], NoReturn]) -> None:
    pass


expects_callable(Foo())         # No error
expects_callable(FooChild())    # No error
expects_callable(Bad())         # Error here: Bad.__call__ has an incompatible signature

基本上，如果一个类有一个 __call__ 方法，则隐含地假定该类也是一个可调用的.

Basically, if a class has a __call__ method, it's implicitly assumed that class is also a callable.

这篇关于Callable 是无效的基类?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！