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How does python know when to pass in `self`?

Consider a trivial example:
class C:
#staticmethod
def my_static_method():
print("static")
def my_instance_method(self):
print("self")
When I call C().my_static_method(), python doesn't pass the instance of C into my_static_method, and the descriptor that my_static_method references doesn't expect an instance of C, either.
This makes sense.
But then when I call C().my_instance_method(), how does python know to pass the instance of C that I'm calling my_instance_method from in as an argument, without me specifying anything?

As the link explains, function objects are descriptors! Just like staticmethod objects.
They have a __get__ method which returns a bound-method object, which essentially just partially applies the instance itself as the first positional argument. Consider:
>>> def foo(self):
... return self.bar
...
>>> class Baz:
... bar = 42
...
>>> baz = Baz()
>>> bound_method = foo.__get__(baz, Baz)
>>> bound_method
<bound method foo of <__main__.Baz object at 0x7ffcd001c7f0>>
>>> method()
42

By adding the #staticmethod decorator to my_static_method, you told python not to pass the calling instance of C into the function. So you can call this function as C.my_static_method().
By calling C() you created an instance of C. Then you called the non static function my_instance_method() which Python happily passed your new instance of C as the first parameter.
What happens when you call C.my_instance_method() ?
Rhetorical: You'll get a "missing one required arg self" exception -- since my_instance_method only works when calling from an instance unless you decorate it as static.
Of course you can still call the static member from an instance C().my_static_method() but you don't have a self param so no access to the instance.

The key point here is that methods are just functions that happen to be attributes of a class. The actual magic, in Python, happens in the attribute lookup process. The link you give explains earlier just how much happens every time x.y happens in Python. (Remember, everything is an object; that includes functions, classes, modules, type (which is an instance of itself)...)
This process is why descriptors can work at all; why we need explicit self; and why we can do fun things like calling a method with normal function call syntax (as long as we look it up from the class rather than an instance), alias it, mimic the method binding process with functools.partial....
Suppose we have c = C(). When you do c.my_instance_method (never mind calling it for now), Python looks for my_instance_method in type(c) (i.e., in the C class), and also checks if it's a descriptor, and also if it's specifically a data descriptor. Functions are non-data descriptors; even outside of a class, you can write
>>> def x(spam): return spam
...
>>> x.__get__
<method-wrapper '__get__' of function object at 0x...>
Because of the priority rules, as long as c doesn't directly have an attribute attached with the same name, the function will be found in C and its __get__ will be used. Note that the __get__ in question comes from the class - but it isn't using the same process as x.__get__ above. That code looks in the class because that's one of the places checked for an attribute lookup; but when c.my_instance_method redirects to C.my_instance_method.__get__, it's looking there directly - attaching a __get__ attribute directly to the function wouldn't change anything (which is why staticmethod is implemented as a class instead).
That __get__ implements the actual method binding. Let's pretend we found x as a method in the str class:
>>> x.__get__('spam', str)
<bound method x of 'spam'>
>>> x.__get__('spam', str)()
'spam'
Remember, although the function in question takes three arguments, we're calling __get__, itself, as a method - so x gets bound to it in the same way. Equivalently, and more faithful to the actual process:
>>> type(x).__get__(x, 'spam', str)
<bound method x of 'spam'>
>>> type(x).__get__(x, 'spam', str)()
'spam'
So what exactly is that "bound method", anyway?
>>> bound = type(x).__get__(x, 'spam', str)
>>> type(bound)
<class 'method'>
>>> bound.__call__
<method-wrapper '__call__' of method object at 0x...>
>>> bound.__func__
<function x at 0x...>
>>> bound.__self__
'spam'
>>> type(bound)(x, 'eggs')
<bound method x of 'eggs'>
Pretty much what you'd expect: it's a callable object that stores and uses the original function and self value, and does the obvious thing in __call__.

Using `super()` within `__init_subclass__` doesn't find parent's classmethod [duplicate]

This question already has an answer here:
Why does a classmethod's super need a second argument?
(1 answer)
Closed 4 years ago.
I try to access the classmethod of a parent from within __init_subclass__ however that doesn't seem to work.
Suppose the following example code:
class Foo:
def __init_subclass__(cls):
print('init', cls, cls.__mro__)
super(cls).foo()
#classmethod
def foo(cls):
print('foo')
class Bar(Foo):
pass
which produces the following exception:
AttributeError: 'super' object has no attribute 'foo'
The cls.__mro__ however shows that Foo is a part of it: (<class '__main__.Bar'>, <class '__main__.Foo'>, <class 'object'>).
So I don't understand why super(cls).foo() doesn't dispatch to Foo.foo. Can someone explain this?

A normal super object (what you normally get from calling super(MyType, self) or super() or super(MyType, myobj)) keeps track of both the type and the object it was created with. Whenever you look up an attribute on the super, it skips over MyType in the method resolution order, but if it finds a method it binds it to that self object.
An unbound super has no self object. So, super(cls) skips over cls in the MRO to find the method foo, and then binds it to… oops, it has nothing to call it on.
So, what things can you call a classmethod on? The class itself, or a subclass of it, or an instance of that class or subclass. So, any of those will work as the second argument to super here, the most obvious one being:
super(cls, cls)
This is somewhat similar to the difference between staticmethods (bound staticmethods are actually bound to nothing) and classmethods (bound classmethods are bound to the class instead of an instance), but it's not quite that simple.
If you want to know why an unbound super doesn't work, you have to understand what an unbound super really is. Unfortunately, the only explanation in the docs is:
If the second argument is omitted, the super object returned is unbound.
What does this mean? Well, you can try to work it out from first principles as a parallel to what it means for a method to be unbound (except, of course, that unbound methods aren't a thing in modern Python), or you can read the C source, or the original introduction to 2.2's class-type unification (including a pure-Python super clone).
A super object has a __self__ attribute, just like a method object. And super(cls) is missing its __self__, just like str.split is.1
You can't use an unbound super explicitly the way you can with an unbound method (e.g., str.split('123', '2') does the same as '123'.split('2'), but super(cls).foo(cls) doesn't work the same as super(cls, cls).foo()). But you can use them implicitly, the same way you do with unbound methods all the time without normally thinking about it.
If you don't know how methods work, the tl'dr is: when you evaluate myobj.mymeth, Python looks up mymeth, doesn't find it on myobj itself, but does find it on the type, so it checks whether it's a non-data descriptor, and, if so, calls its __get__ method to bind it to myobj.
So, unbound methods2 are non-data descriptors whose __get__ method returns a bound method. Unbound #classmethods are similar, but their __get__ ignores the object and returns a bound method bound to the class. And so on.
And unbound supers are non-data descriptors whose __get__ method returns a bound super.
Example (credit to wim for coming up with the closest thing to a use for unbound super that I've seen):
class A:
def f(self): print('A.f')
class B(A):
def f(self): print('B.f')
b = B()
bs = super(B)
B.bs = bs
b.bs.f()
We created an unbound super bs, stuck it on the type B, and then b.bs is a normal bound super, so b.bs.f is A.f, just like super().f would have been inside a B method.
Why would you want to do that? I'm not sure. I've written all kinds of ridiculously dynamic and reflective code in Python (e.g., for transparent proxies to other interpreters), and I can't remember ever needing an unbound super. But if you ever need it, it's there.
1. I'm cheating a bit here. First, unbound methods aren't a thing anymore in Python 3—but functions work the same way, so Python uses them where it used to use unbound methods. Second, str.split, being a C builtin, wasn't properly an unbound method even in 2.x—but it acts like one anyway, at least as far as we're concerned here.
2. Actually plain-old functions.

Are there more than three types of methods in Python?

I understand there are at least 3 kinds of methods in Python having different first arguments:
instance method - instance, i.e. self
class method - class, i.e. cls
static method - nothing
These classic methods are implemented in the Test class below including an usual method:
class Test():
def __init__(self):
pass
def instance_mthd(self):
print("Instance method.")
#classmethod
def class_mthd(cls):
print("Class method.")
#staticmethod
def static_mthd():
print("Static method.")
def unknown_mthd():
# No decoration --> instance method, but
# No self (or cls) --> static method, so ... (?)
print("Unknown method.")
In Python 3, the unknown_mthd can be called safely, yet it raises an error in Python 2:
>>> t = Test()
>>> # Python 3
>>> t.instance_mthd()
>>> Test.class_mthd()
>>> t.static_mthd()
>>> Test.unknown_mthd()
Instance method.
Class method.
Static method.
Unknown method.
>>> # Python 2
>>> Test.unknown_mthd()
TypeError: unbound method unknown_mthd() must be called with Test instance as first argument (got nothing instead)
This error suggests such a method was not intended in Python 2. Perhaps its allowance now is due to the elimination of unbound methods in Python 3 (REF 001). Moreover, unknown_mthd does not accept args, and it can be bound to called by a class like a staticmethod, Test.unknown_mthd(). However, it is not an explicit staticmethod (no decorator).
Questions
Was making a method this way (without args while not explicitly decorated as staticmethods) intentional in Python 3's design? UPDATED
Among the classic method types, what type of method is unknown_mthd?
Why can unknown_mthd be called by the class without passing an argument?
Some preliminary inspection yields inconclusive results:
>>> # Types
>>> print("i", type(t.instance_mthd))
>>> print("c", type(Test.class_mthd))
>>> print("s", type(t.static_mthd))
>>> print("u", type(Test.unknown_mthd))
>>> print()
>>> # __dict__ Types, REF 002
>>> print("i", type(t.__class__.__dict__["instance_mthd"]))
>>> print("c", type(t.__class__.__dict__["class_mthd"]))
>>> print("s", type(t.__class__.__dict__["static_mthd"]))
>>> print("u", type(t.__class__.__dict__["unknown_mthd"]))
>>> print()
i <class 'method'>
c <class 'method'>
s <class 'function'>
u <class 'function'>
i <class 'function'>
c <class 'classmethod'>
s <class 'staticmethod'>
u <class 'function'>
The first set of type inspections suggests unknown_mthd is something similar to a staticmethod. The second suggests it resembles an instance method. I'm not sure what this method is or why it should be used over the classic ones. I would appreciate some advice on how to inspect and understand it better. Thanks.
REF 001: What's New in Python 3: “unbound methods” has been removed
REF 002: How to distinguish an instance method, a class method, a static method or a function in Python 3?
REF 003: What's the point of #staticmethod in Python?

Some background: In Python 2, "regular" instance methods could give rise to two kinds of method objects, depending on whether you accessed them via an instance or the class. If you did inst.meth (where inst is an instance of the class), you got a bound method object, which keeps track of which instance it is attached to, and passes it as self. If you did Class.meth (where Class is the class), you got an unbound method object, which had no fixed value of self, but still did a check to make sure a self of the appropriate class was passed when you called it.
In Python 3, unbound methods were removed. Doing Class.meth now just gives you the "plain" function object, with no argument checking at all.
Was making a method this way intentional in Python 3's design?
If you mean, was removal of unbound methods intentional, the answer is yes. You can see discussion from Guido on the mailing list. Basically it was decided that unbound methods add complexity for little gain.
Among the classic method types, what type of method is unknown_mthd?
It is an instance method, but a broken one. When you access it, a bound method object is created, but since it accepts no arguments, it's unable to accept the self argument and can't be successfully called.
Why can unknown_mthd be called by the class without passing an argument?
In Python 3, unbound methods were removed, so Test.unkown_mthd is just a plain function. No wrapping takes place to handle the self argument, so you can call it as a plain function that accepts no arguments. In Python 2, Test.unknown_mthd is an unbound method object, which has a check that enforces passing a self argument of the appropriate class; since, again, the method accepts no arguments, this check fails.

#BrenBarn did a great job answering your question. This answer however, adds a plethora of details:
First of all, this change in bound and unbound method is version-specific, and it doesn't relate to new-style or classic classes:
2.X classic classes by default
>>> class A:
... def meth(self): pass
...
>>> A.meth
<unbound method A.meth>
>>> class A(object):
... def meth(self): pass
...
>>> A.meth
<unbound method A.meth>
3.X new-style classes by default
>>> class A:
... def meth(self): pass
...
>>> A.meth
<function A.meth at 0x7efd07ea0a60>
You've already mentioned this in your question, it doesn't hurt to mention it twice as a reminder.
>>> # Python 2
>>> Test.unknown_mthd()
TypeError: unbound method unknown_mthd() must be called with Test instance as first argument (got nothing instead)
Moreover, unknown_mthd does not accept args, and it can be bound to a class like a staticmethod, Test.unknown_mthd(). However, it is not an explicit staticmethod (no decorator)
unknown_meth doesn't accept args, normally because you've defined the function without so that it does not take any parameter. Be careful and cautious, static methods as well as your coded unknown_meth method will not be magically bound to a class when you reference them through the class name (e.g, Test.unknown_meth). Under Python 3.X Test.unknow_meth returns a simple function object in 3.X, not a method bound to a class.
1 - Was making a method this way (without args while not explicitly decorated as staticmethods) intentional in Python 3's design? UPDATED
I cannot speak for CPython developers nor do I claim to be their representative, but from my experience as a Python programmer, it seems like they wanted to get rid of a bad restriction, especially given the fact that Python is extremely dynamic, not a language of restrictions; why would you test the type of objects passed to class methods and hence restrict the method to specific instances of classes? Type testing eliminates polymorphism. It would be decent if you just return a simple function when a method is fetched through the class which functionally behaves like a static method, you can think of unknown_meth to be static method under 3.X so long as you're careful not to fetch it through an instance of Test you're good to go.
2- Among the classic method types, what type of method is unknown_mthd?
Under 3.X:
>>> from types import *
>>> class Test:
... def unknown_mthd(): pass
...
>>> type(Test.unknown_mthd) is FunctionType
True
It's simply a function in 3.X as you could see. Continuing the previous session under 2.X:
>>> type(Test.__dict__['unknown_mthd']) is FunctionType
True
>>> type(Test.unknown_mthd) is MethodType
True
unknown_mthd is a simple function that lives inside Test__dict__, really just a simple function which lives inside the namespace dictionary of Test. Then, when does it become an instance of MethodType? Well, it becomes an instance of MethodType when you fetch the method attribute either from the class itself which returns an unbound method or its instances which returns a bound method. In 3.X, Test.unknown_mthd is a simple function--instance of FunctionType, and Test().unknown_mthd is an instance of MethodType that retains the original instance of class Test and adds it as the first argument implicitly on function calls.
3- Why can unknown_mthd be called by the class without passing an argument?
Again, because Test.unknown_mthd is just a simple function under 3.X. Whereas in 2.X, unknown_mthd not a simple function and must be called be passed an instance of Test when called.

Are there more than three types of methods in Python?
Yes. There are the three built-in kinds that you mention (instance method, class method, static method), four if you count #property, and anyone can define new method types.
Once you understand the mechanism for doing this, it's easy to explain why unknown_mthd is callable from the class in Python 3.
A new kind of method
Suppose we wanted to create a new type of method, call it optionalselfmethod so that we could do something like this:
class Test(object):
#optionalselfmethod
def optionalself_mthd(self, *args):
print('Optional-Self Method:', self, *args)
The usage is like this:
In [3]: Test.optionalself_mthd(1, 2)
Optional-Self Method: None 1 2
In [4]: x = Test()
In [5]: x.optionalself_mthd(1, 2)
Optional-Self Method: <test.Test object at 0x7fe80049d748> 1 2
In [6]: Test.instance_mthd(1, 2)
Instance method: 1 2
optionalselfmethod works like a normal instance method when called on an instance, but when called on the class, it always receives None for the first parameter. If it were a normal instance method, you would always have to pass an explicit value for the self parameter in order for it to work.
So how does this work? How you can you create a new method type like this?
The Descriptor Protocol
When Python looks up a field of an instance, i.e. when you do x.whatever, it check in several places. It checks the instance's __dict__ of course, but it also checks the __dict__ of the object's class, and base classes thereof. In the instance dict, Python is just looking for the value, so if x.__dict__['whatever'] exists, that's the value. However, in the class dict, Python is looking for an object which implements the Descriptor Protocol.
The Descriptor Protocol is how all three built-in kinds of methods work, it's how #property works, and it's how our special optionalselfmethod will work.
Basically, if the class dict has a value with the correct name1, Python checks if it has an __get__ method, and calls it like type(x).whatever.__get__(x, type(x)) Then, the value returned from __get__ is used as the field value.
So for example, a trivial descriptor which always returns 3:
class GetExample:
def __get__(self, instance, cls):
print("__get__", instance, cls)
return 3
class Test:
get_test = GetExample()
Usage is like this:
In[22]: x = Test()
In[23]: x.get_test
__get__ <__main__.Test object at 0x7fe8003fc470> <class '__main__.Test'>
Out[23]: 3
Notice that the descriptor is called with both the instance and the class type. It can also be used on the class:
In [29]: Test.get_test
__get__ None <class '__main__.Test'>
Out[29]: 3
When a descriptor is used on a class rather than an instance, the __get__ method gets None for self, but still gets the class argument.
This allows a simple implementation of methods: functions simply implement the descriptor protocol. When you call __get__ on a function, it returns a bound method of instance. If the instance is None, it returns the original function. You can actually call __get__ yourself to see this:
In [30]: x = object()
In [31]: def test(self, *args):
...: print(f'Not really a method: self<{self}>, args: {args}')
...:
In [32]: test
Out[32]: <function __main__.test>
In [33]: test.__get__(None, object)
Out[33]: <function __main__.test>
In [34]: test.__get__(x, object)
Out[34]: <bound method test of <object object at 0x7fe7ff92d890>>
#classmethod and #staticmethod are similar. These decorators create proxy objects with __get__ methods which provide different binding. Class method's __get__ binds the method to the instance, and static method's __get__ doesn't bind to anything, even when called on an instance.
The Optional-Self Method Implementation
We can do something similar to create a new method which optionally binds to an instance.
import functools
class optionalselfmethod:
def __init__(self, function):
self.function = function
functools.update_wrapper(self, function)
def __get__(self, instance, cls):
return boundoptionalselfmethod(self.function, instance)
class boundoptionalselfmethod:
def __init__(self, function, instance):
self.function = function
self.instance = instance
functools.update_wrapper(self, function)
def __call__(self, *args, **kwargs):
return self.function(self.instance, *args, **kwargs)
def __repr__(self):
return f'<bound optionalselfmethod {self.__name__} of {self.instance}>'
When you decorate a function with optionalselfmethod, the function is replaced with our proxy. This proxy saves the original method and supplies a __get__ method which returns a boudnoptionalselfmethod. When we create a boundoptionalselfmethod, we tell it both the function to call and the value to pass as self. Finally, calling the boundoptionalselfmethod calls the original function, but with the instance or None inserted into the first parameter.
Specific Questions
Was making a method this way (without args while not explicitly
decorated as staticmethods) intentional in Python 3's design? UPDATED
I believe this was intentional; however the intent would have been to eliminate unbound methods. In both Python 2 and Python 3, def always creates a function (you can see this by checking a type's __dict__: even though Test.instance_mthd comes back as <unbound method Test.instance_mthd>, Test.__dict__['instance_mthd'] is still <function instance_mthd at 0x...>).
In Python 2, function's __get__ method always returns a instancemethod, even when accessed through the class. When accessed through an instance, the method is bound to that instance. When accessed through the class, the method is unbound, and includes a mechanism which checks that the first argument is an instance of the correct class.
In Python 3, function's __get__ method will return the original function unchanged when accessed through the class, and a method when accessed through the instance.
I don't know the exact rationale but I would guess that type-checking of the first argument to a class-level function was deemed unnecessary, maybe even harmful; Python allows duck-typing after all.
Among the classic method types, what type of method is unknown_mthd?
unknown_mthd is a plain function, just like any normal instance method. It only fails when called through the instance because when method.__call__ attempts to call the function unknown_mthd with the bound instance, it doesn't accept enough parameters to receive the instance argument.
Why can unknown_mthd be called by the class without passing an
argument?
Because it's just a plain function, same as any other function. I just doesn't take enough arguments to work correctly when used as an instance method.
You may note that both classmethod and staticmethod work the same whether they're called through an instance or a class, whereas unknown_mthd will only work correctly when when called through the class and fail when called through an instance.
1. If a particular name has both a value in the instance dict and a descriptor in the class dict, which one is used depends on what kind of descriptor it is. If the descriptor only defines __get__, the value in the instance dict is used. If the descriptor also defines __set__, then it's a data-descriptor and the descriptor always wins. This is why you can assign over top of a method but not a #property; method only define __get__, so you can put things in the same-named slot in the instance dict, while #properties define __set__, so even if they're read-only, you'll never get a value from the instance __dict__ even if you've previously bypassed property lookup and stuck a value in the dict with e.g. x.__dict__['whatever'] = 3.

How do Python tell “this is called as a function”?

A callable object is supposed to be so by defining __call__. A class is supposed to be an object… or at least with some exceptions. This exception is what I'm failing to formally clarify, thus this question posted here.
Let A be a simple class:
class A(object):
def call(*args):
return "In `call`"
def __call__(*args):
return "In `__call__`"
The first function is purposely named “call”, to make clear the purpose is the comparison with the other.
Let's instantiate it and forget about the expression it implies:
a = A() # Think of it as `a = magic` and forget about `A()`
Now what's worth:
print(A.call())
print(a.call())
print(A())
print(a())
Result in:
>>> In `call`
>>> In `call`
>>> <__main__.A object at 0xNNNNNNNN>
>>> In `__call__`
The output (third statement not running __call__) does not come as a surprise, but when I think every where it is said “Python class are objects”…
This, more explicit, however run __call__
print(A.__call__())
print(a.__call__())
>>> “In `__call__`”
>>> “In `__call__`”
All of this is just to show how finally A() may looks strange.
There are exception in Python rules, but the documentation about “object.call” does not say a lot about __call__… not more than that:
3.3.5. Emulating callable objects
object.__call__(self[, args...])
Called when the instance is “called” as a function; […]
But how do Python tell “it's called as a function” and honour or not the object.__call__ rule?
This could be a matter of type, but even type has object as its base class.
Where can I learn more (and formally) about it?
By the way, is there any difference here between Python 2 and Python 3?
----- %< ----- edit ----- >% -----
Conclusions and other experiments after one answer and one comment
Update #1
After #Veedrac's answer and #chepner's comment, I came to this other test, which complete the comments from both:
class M(type):
def __call__(*args):
return "In `M.__call__`"
class A(object, metaclass=M):
def call(*args):
return "In `call`"
def __call__(*args):
return "In `A.__call__`"
print(A())
The result is:
>>> In `M.__call__`
So it seems that's the meta‑class which drives the “call” operations. If I understand correctly, the meta‑class does not matter only with class, but also with classes instances.
Update #2
Another relevant test, which shows this is not an attribute of the object which matters, but an attribute of the type of the object:
class A(object):
def __call__(*args):
return "In `A.__call__`"
def call2(*args):
return "In `call2`"
a = A()
print(a())
As expected, it prints:
>>> In `A.__call__`
Now this:
a.__call__ = call2
print(a())
It prints:
>>> In `A.__call__`
The same a before the attribute was assigned. It does not print In call2, it's still In A.__call__. That's important to note and also explain why that's the __call__ of the meta‑class which was invoked (keep in mind the meta‑class is the type of the class object). The __call__ used to call as function, is not from the object, it's from its type.

x(*args, **kwargs) is the same as type(x).__call__(x, *args, **kwargs).
So you have
>>> type(A).__call__(A)
<__main__.A object at 0x7f4d88245b50>
and it all makes sense.
chepner points out in the comments that type(A) == type. This is kind-of wierd, because type(A)(A) just gives type again! But remember that we're instead using type(A).__call__(A) which is not the same.
So this resolves to type.__call__(A). This is the constructor function for classes, which builds the data-structures and does all the construction magic.
The same is true of most dunder (double underscore) methods, such as __eq__. This is partially an optimisation in those cases.

what is super(type) in python?

Class definition:
class A(object):
def foo(self):
print "A"
class B(object):
def foo(self):
print "B"
class C(A, B):
def foo(self):
print "C"
Output:
>>> super(C)
<super: <class 'C'>, NULL>
>>> super(C).foo
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
AttributeError: 'super' object has no attribute 'foo'
What is the use of super(type) if we can't access attributes of a class?

super(type) is an "unbound" super object. The docs on super discuss that, but don't really elaborate what an "unbound" super object is or does. It is simply a fact of the language that you cannot use them in the manner you are attempting to use them.
This is perhaps what you want:
>>> super(C, C).foo is B.foo
True
That said, what good is an unbound super object? I had to look this up, myself, and found a decent answer here. Note however, that the article's conclusion is that unbound super is a language wart, has no practical use, and should be removed from the language (and having read the article, I agree). The article's explanation on unbound super starts with:
Unbound super objects must be turned into bound objects in order to make them to dispatch properly. That can be done via the descriptor protocol. For instance, I can convert super(C1) in a super object bound to c1 in this way:
>>> c1 = C1()
>>> boundsuper = super(C1).__get__(c1, C1) # this is the same as super(C1, c1)
So, that doesn't seem useful, but the article goes on:
Having established that the unbound syntax does not return unbound methods one might ask what its purpose is. The answer is that super(C) is intended to be used as an attribute in other classes. Then the descriptor magic will automatically convert the unbound syntax in the bound syntax. For instance:
>>> class B(object):
... a = 1
>>> class C(B):
... pass
>>> class D(C):
... sup = super(C)
>>> d = D()
>>> d.sup.a
1
This works since d.sup.a calls super(C).__get__(d,D).a which is turned into super(C, d).a and retrieves B.a.
There is a single use case for the single argument syntax of super that I am aware of, but I think it gives more troubles than advantages. The use case is the implementation of autosuper made by Guido on his essay about new-style classes.
The idea there is to use the unbound super objects as private attributes. For instance, in our example, we could define the private attribute __sup in the class C as the unbound super object super(C):
>>> C._C__sup = super(C)
But do note that the article continues to describe the problems with this (it doesn't quite work correctly, I think mostly due to the fact that the MRO is dependent on the class of the instance you are dealing with, and thus given an instance, some class X's superclass may be different depending on the instance of X we are given).

To accomplish what you want, you need to call it like this:
>>> myC = C()
>>> super(C,myC).foo()
A

Note that there is a NULL reference to some object. It basically needs a class and an instance of a related object for it to function.
>>> super(C, C()).foo
<bound method C.foo of <__main__.C object at 0x225dc50>>
See: Understanding Python super() with __init__() methods for more details.