super() fails with error: TypeError "argument 1 must be type, not classobj" when parent does not inherit from object

Question

I get some error that I can't figure out. Any clue what is wrong with my sample code?

class B:
    def meth(self, arg):
        print arg

class C(B):
    def meth(self, arg):
        super(C, self).meth(arg)

print C().meth(1)

I got the sample test code from help of 'super' built-in method.

Here is the error:

Traceback (most recent call last):
  File "./test.py", line 10, in ?
    print C().meth(1)
  File "./test.py", line 8, in meth
    super(C, self).meth(arg)
TypeError: super() argument 1 must be type, not classobj

FYI, here is the help(super) from python itself:

Help on class super in module __builtin__:

class super(object)
 |  super(type) -> unbound super object
 |  super(type, obj) -> bound super object; requires isinstance(obj, type)
 |  super(type, type2) -> bound super object; requires issubclass(type2, type)
 |  Typical use to call a cooperative superclass method:
 |  class C(B):
 |      def meth(self, arg):
 |          super(C, self).meth(arg)
 |

Answer 1

Your problem is that class B is not declared as a "new-style" class. Change it like so:

class B(object):

and it will work.

super() and all subclass/superclass stuff only works with new-style classes. I recommend you get in the habit of always typing that (object) on any class definition to make sure it is a new-style class.

Old-style classes (also known as "classic" classes) are always of type classobj; new-style classes are of type type. This is why you got the error message you saw:

TypeError: super() argument 1 must be type, not classobj

Try this to see for yourself:

class OldStyle:
    pass

class NewStyle(object):
    pass

print type(OldStyle)  # prints: <type 'classobj'>

print type(NewStyle) # prints <type 'type'>

Note that in Python 3.x, all classes are new-style. You can still use the syntax from the old-style classes but you get a new-style class. So, in Python 3.x you won't have this problem.