Up to Python 2.1, old-style classes were the only flavour available to the user.
The concept of (old-style) class is unrelated to the concept of type:
x is an instance of an old-style class, then
designates the class of
type(x) is always
This reflects the fact that all old-style instances, independently of
their class, are implemented with a single built-in type, called
New-style classes were introduced in Python 2.2 to unify the concepts of class and type.
A new-style class is simply a user-defined type, no more, no less.
If x is an instance of a new-style class, then
type(x) is typically
the same as
x.__class__ (although this is not guaranteed – a
new-style class instance is permitted to override the value returned
The major motivation for introducing new-style classes is to provide a unified object model with a full meta-model.
It also has a number of immediate benefits, like the ability to
subclass most built-in types, or the introduction of "descriptors",
which enable computed properties.
For compatibility reasons, classes are still old-style by default.
New-style classes are created by specifying another new-style class
(i.e. a type) as a parent class, or the "top-level type" object if no
other parent is needed.
The behaviour of new-style classes differs from that of old-style
classes in a number of important details in addition to what type
Some of these changes are fundamental to the new object model, like
the way special methods are invoked. Others are "fixes" that could not
be implemented before for compatibility concerns, like the method
resolution order in case of multiple inheritance.
Python 3 only has new-style classes.
No matter if you subclass from
object or not, classes are new-style
in Python 3.
Maybe it would be better if you explicit what you want to do. Why do you want to create an unnamed class? Does it conform to an interface? Unnamed classes are quite limited, they cannot be used as parameters to functions, they cannot be used as template type-parameters...
Now if you are implmenting an interface then you can pass references to that interface:
NOTE: For all those answers that consider template arguments as an option, unnamed classes cannot be passed as template type arguments.
C++ Standard. 14.3.1, paragraph 2:
If you test with comeau compiler (the link is for the online tryout) you will get the following error:
As a side note, comeau compiler is the most standard compliant compiler I know of, besides being the one with the most helpful error diagnostics I have tried.
NOTE: Comeau and gcc (g++ 4.0) give an error with the code above. Intel compiler (and from other peoples comments MSVS 2008) accept using unnamed classes as template parameters, against the standard.