学习abc模块时为何空类被判定为Hashable抽象基类的实例？

Great question! Let's break down why your code returns True step by step.

1. How abc.Hashable checks for membership #

The abc.Hashable abstract base class (ABC) doesn't rely solely on explicit inheritance to recognize a type as hashable. Instead, it uses a __subclasshook__ method to perform a dynamic check: it verifies if the target class has a valid __hash__ method.

2. Default behavior of user-defined classes #

When you define an empty class like foo without overriding any methods:

• It inherits all default methods from object, including __hash__ and __eq__.
• Since you haven't overridden __eq__, Python keeps the default __hash__ method intact. (Note: If you did override __eq__ without providing a custom __hash__, Python would automatically set __hash__ = None for the class.)

3. Applying this to your code #

When you run isinstance(foo, abc.Hashable):

• The check isn't targeting instances of foo—it's examining the class object foo itself.
• The foo class retains the valid __hash__ method inherited from object, so abc.Hashable's subclass hook recognizes it as a hashable type.
• That's why the result is True.

Let's test this with a modification #

If you override __eq__ without defining __hash__, the result will change:

from collections import abc
class foo:
    def __eq__(self, other):
        return isinstance(other, foo)
print(isinstance(foo, abc.Hashable)) # False

Here, Python sets foo.__hash__ = None because we defined __eq__ but not __hash__, so abc.Hashable no longer recognizes the class as hashable.

Summary #

The core point is that abc.Hashable uses structural subtyping (checking for the presence of required methods) rather than just explicit inheritance. Your empty class keeps the default __hash__ method from object, so it passes the hashable check.

内容的提问来源于stack exchange，提问作者yoyoyoyo123