How is the reference of mixin pass to another mixin in python?

Question

If a class is inheriting multiple class, why can the superclasses access functions of the other superclasses mutually? Where is the superclass getting the referencing from?

For example

class A():
    def a_method(self):
        print "I am a"

class B():
    def b_method(self):
        self.a_method()

class test(A, B):
    def test_method(self):
        self.b_method()

if __name__ == "__main__":
    test_instance = test()
    # Will print a_method
    test_instance.test_method()

    test_b = B()

    try:
        # will raise exception
        test_b.b_method()
    except Exception as e:
        print e

Answer 1

When you define a class as inheriting from two superclasses

class test(A, B):

it inherits the methods from both superclasses into the same namespace. So, from test() , you can call both self.a_method() and self.b_method() . Your question, I assume, is why calling self.b_method() works from an instance of test , but not an instance of B . It works in test because both methods are in the same namespace, and when b_method() calls a_method() , it can be "seen" from inside the class, and the call succeeds. When instantiating B , which does not inherit from A , a_method() is not visible, and so an exception is raised.

The methods and attributes associated with a class or instance can be examined with dir :

>>> test_instance = test()
>>> dir(test_instance)
['__doc__', '__module__', 'a_method', 'b_method', 'test_method']