在python中，为什么super具有多重继承行为？

Question

So we have a code snippet below. I don't understand why it behave this way. Why does super(B, self).go() resolves to the go method of the C class?

class A(object):
    def go(self):
        print("go A go!")

class B(A):
    def go(self):
        super(B, self).go()
        print("go B go!")

class C(A):
    def go(self):
        super(C, self).go()
        print("go C go!")

class D(B, C):
    def go(self):
        super(D, self).go()
        print("go D go!")

d = D()
d.go()
# go A go!
# go C go!
# go B go!
# go D go!

Answer 1

Despite its name, super does not necessarily refer to a superclass. super(B, self) refers to the class in self 's MRO that follows B .

You can see the MRO for D with

>>> D.__mro__
(<class '__main__.D'>, <class '__main__.B'>, <class '__main__.C'>, <class '__main__.A'>, <type 'object'>)

This means that from d.go , super(D, self) refers to B . When B.go is called as a result, super(B, self) refers to C , not A . This is because self is still an instance of D , so it is D.__mro__ that dictates what gets called next, not the static superclass of B .

The most important thing to remember about super is that inside Foo.go , you do not know what class super(Foo, self) will refer to, because you do not know if self is an instance of Foo or a descendent of Foo .