`type.__instancecheck__` 是否实现了 `isinstance` 的默认行为？

Question

The docs say the following:

[...] If defined, called to implement isinstance(instance, class) .

The wording suggests that it makes a difference whether __instancecheck__ is defined or not. Ie it doesn't say "Override to implement isinstance " where it would be clear that the base implementation ( type.__instancecheck__ ) provides the default behavior.

To be more specific I would like to know whether the following meta class results in similar behavior as if I omitted the definition of __instancecheck__ :

class Meta(type):
    def __instancecheck__(self, instance):
        return super().__instancecheck__(instance)

class Test(metaclass=Meta):
    pass

isinstance(<something>, Test)

My intuition says yes, but the wording of the docs makes me worry that isinstance takes a different path based on whether __instancecheck__ is defined or not. In the end I would like to know whether I can fall back to super().__instancecheck__(instance) in my own implementation of __instancecheck__ in order to get the default behavior.

Answer 1

isinstance doesn't act any different with or without __instancecheck__ being defined. Also, "falling back" on super().__instancecheck__(instance) is unnecessary. According to https://bugs.python.org/issue35083 , "any object x is always considered to be an instance of type(x) , and this cannot be overridden."