can someone explain 'super' to me?

[Lie Ryan]

On 12/5/2009 9:27 PM, Michael wrote:
> It seems like it can return either a class or an instance of a class.
> Like
> super( C, self)
> is like casting self as superclass C.
> However if you omit the second argument entirely you get a class.

Inside a class C: these are all equivalent:
super().method(arg) # python 3
super(C, self).method(arg)
super(C).method(self, arg)

it is similar to how you can call
class C(object):
     def method(self, arg):
         pass

inst = C()
# these are equivalent
inst.method(arg)
C.method(inst, arg)


python 2.x restricts the first argument of an unbound method to instance 
of the class; python 3.x does not have such restriction. Thus, it is 
possible in python 3.x to have:
 >>> class A(object):
...     pass
...
 >>> class B(object):
...     def anom(self):
...         print(self)
...
 >>> a = A()
 >>> B.anom(a)
<__main__.A object at 0x0165C630>

the same thing in python 2 would be an error.

> The former is considered a "bound" object. I'm really not clear on the
> idea of "binding" in Python.

The first argument of a bound method (the argument self) is 
automatically redirected to the instance. Notice when you called a method:
class A(object):
     # this declaration have 3 arguments
     def foo(self, a, b):
         pass

a = A()
# called by only 2 arguments
a.foo(1, 2)

because a.foo binds method A.foo() and `a`; and `a` is passed implicitly 
as the first argument to A.foo(a, 1, 2).