call super in subclass created by type

Question

Update: made some progress and simplified the example, although I'm currently stumped by the latest error. I have no idea why the super().__init__ method is unbound.

import types

class VeryImportantSuperClass(object):

  def __init__(self, anArg, anotherArg):
    self.anArg = anArg
    #Extremely clever code here

def createSubclassAttempt1(name):
  source = 'def __init__(self, arg):\n' +\
           '  super(' + name + ', self).__init__(arg, 6.02e23)\n' +\
           '  self.foo = self.anArg + 3\n'
  cls = type(name, (VeryImportantSuperClass,), {})
  d = {name: cls}
  exec source in d
  cls.__init__ = types.MethodType(d['__init__'], cls)
  return cls

if __name__ == '__main__':
  cls = createSubclassAttempt1('Foo')
  cls('foo')

output:

Traceback (most recent call last):
  File "/home/newb/eclipseWorkspace/TestArea/subclassWithType.py", line 27, in <module>
    cls('foo')
  File "<string>", line 2, in __init__
TypeError: unbound method __init__() must be called with Foo instance as first argument (got str instance instead)

There's got to be some way to call a superclass method from a subclass created by type, but I'm dashed if I can see how.

Answer 1

This seems to work and do what you want. Theself.foo = self.anArg + 3statement had to be changed toself.foo = self.anArg + "3"to avoid aTypeErrorwhen attempting to concatenate astrand anint objects which is what happens with acls('f00')call as shown in the code in your question.

import types

class VeryImportantSuperClass(object):
    def __init__(self, anArg, anotherArg):
        self.anArg = anArg
        #Extremely clever code here

def createSubclassAttempt1(name):
    source = ('def __init__(self, arg):\n'
              '    super(' + name + ', self).__init__(arg, 6.02e23)\n'
              '    self.foo = self.anArg + "3"\n')
    d = {}
    exec(source, d)
    cls = type(name, (VeryImportantSuperClass,), d)
    d[name] = cls
    return cls

if __name__ == '__main__':
    cls = createSubclassAttempt1('Foo')
    inst = cls('foo')
    print(cls('foo').foo)  # prints "foo3"

(Works in both Python 2.7.3 & 3.3)

Answer 2

Use a closure for __init__ in the function creating the class:

class VeryImportantSuperClass(object):
    def __init__(self, anArg, anotherArg):
        self.anArg = anArg

def CreateSubclass(name):
    def sub__init__(self, arg):
        super(klass, self).__init__(arg, 6.02e23)
        self.foo = self.anArg + 3

    klass = type(name, (VeryImportantSuperClass, ),
                 {'__init__': sub__init__})
    return klass


if __name__ == '__main__':
    cls = CreateSubclass('Foo')
    print(cls(3).foo)