python - 继承 - __hash__ 在子类中设置为 None

Question

我设法在 Python 3.4 和 3.7 上重现了这一点。

考虑:

class Comparable:
    def _key(self):
        raise NotImplementedError

    def __hash__(self):
        return hash(self._key())

    def __eq__(self, other):
        ...

    def __lt__(self, other):
        ...


class A(Comparable): pass

class B(A):
    def __str__(self):
        return "d"

    def __eq__(self, other):
        return isinstance(self, type(other))

    def _key(self):
        return str(self),

b = B()

很明显，人们会期望 b.__hash__ 在这里定义，因为它是在 Comparable 下定义的，B 是 B 的子类。

瞧，它已被定义，但计算结果为 None。给了什么？

>> b
<__main__.B object at 0x00000183C9734978>
>> '__hash__' in dir(b)
True
>> b.__hash__

>> b.__hash__ is None
True
>> B.__mro__
(<class '__main__.B'>, <class '__main__.A'>, <class '__main__.Comparable'>, <class 'object'>)
>> isinstance(b, Comparable)
True

如果在 Comparable 和 A 中将 __init__ 实现为 super().__init__()，则会重现相同的行为>.

Answer 1

在 the docs 中找到它:

A class that overrides __eq__() and does not define __hash__() will have its __hash__() implicitly set to None.

和

If a class that overrides __eq__() needs to retain the implementation of __hash__() from a parent class, the interpreter must be told this explicitly by setting __hash__ = <ParentClass>.__hash__

来自工单 1549 :

This was done intentionally -- if you define a comparison without defining a hash, the default hash will not match your comparison, and your objects will misbehave when used as dictionary keys.

(圭多范罗森)