Python's super() function

Question

In the below sample, the last 2 lines in the B.Go() method both call the Go() method from class A. Are they functionally identical? Is the only benefit to using super() that I don't have to know the inherited class name?

class A(object):
    def Go(self):
        print "Calling A.Go()"

class B(A):
    def Go(self):
        super(B, self).Go()
        A.Go(self)

inst = B()
inst.Go()

Answer 1

No, super() does something a direct call to A.Go can't. super(B, self).Go() calls the next method in the method resolution order. It calls the method that would have been called had B not implemented the method at all. In direct linear inheritance this is always A, so there is no real difference (besides the class you repeat.) In the case of multiple inheritance, however, the next class in the normal method resolution order is not necessarily the direct baseclass of the current class. See the explanation for diamond inheritance in Guido's original explanation of these features.

Answer 2

In addition to the excellent answer provided by Thomas Wouters, I'd add that super() is additionally very useful because it is a computed indirection. One key benfit of using super() is that you don't have to specify the delegate class by name, which makes it easy to change the base class later. Raymond Hettinger has a great post on about this: Python's super() considered super!

Answer 3

Not exactly. By using super you need only instantiate B, there is one instance. The second line makes a new instance of A, may possibly have side effects in the initializer, and doesn't share other instance attribute or state (of B instance inst).

Answer 4

They're completely different. Calling A().Go() your contructing a whole new object, that doesn't share any attribute with the instance it was created into.

Your example is too simple, you need to have a state to show the real different behaviour