super and __new__ confusion

Question

As what I just learned, I can use super() this way:
super(class, obj_of_class-or-_subclass_of_class)

Code goes below:

#Case 1 class A(object):     def __init__(self):         print "A init"  class B(A):     def __init__(self):         print "B init"         super(B, self).__init__()  #ok, I can invoke A's __init__ successfully  #Case 2 class A(object):     @classmethod     def foo(cls):         print "A foo"  class B(object):     @classmethod     def foo(cls):         print "B foo"         super(B, cls).foo()   #ok, I can invoke A's foo successfully  #Case 3 class A(object):     def __new__(cls):       print "A new"       return super(A, cls).__new__(cls)  class B(A):     def __new__(cls):       print "B new"       return super(B, cls).__new__()  #Oops, error 

QUESTION:

In case 1 and 2, I can use super successfully without specifying the obj or cls to operate on. But why can't I do the same for __new__? Because, in case 3, if I use super that way, I got an error. But if I use it this way:

super(B, cls).__new__(cls) 

No error.

Answer 1

From the Python release notes on overriding the __new__ method:

__new__ is a static method, not a class method. I initially thought it would have to be a class method, and that's why I added the classmethod primitive. Unfortunately, with class methods, upcalls don't work right in this case, so I had to make it a static method with an explicit class as its first argument.

Since __new__ is a static method, super(...).__new__ returns a static method. There is no binding of cls to the first argument in this case. All arguments need to be supplied explicitly.