How to auto register a class when it's defined

Question

I want to have an instance of class registered when the class is defined. Ideally the code below would do the trick.

registry = {}

def register( cls ):
   registry[cls.__name__] = cls() #problem here
   return cls

@register
class MyClass( Base ):
   def __init__(self):
      super( MyClass, self ).__init__() 

Unfortunately, this code generates the error NameError: global name 'MyClass' is not defined.

What's going on is at the #problem here line I'm trying to instantiate a MyClass but the decorator hasn't returned yet so it doesn't exist.

Is the someway around this using metaclasses or something?

Answer 1

Yes, meta classes can do this. A meta class' __new__ method returns the class, so just register that class before returning it.

class MetaClass(type):
    def __new__(cls, clsname, bases, attrs):
        newclass = super(MetaClass, cls).__new__(cls, clsname, bases, attrs)
        register(newclass)  # here is your register function
        return newclass

class MyClass(object):
    __metaclass__ = MetaClass

The previous example works in Python 2.x. In Python 3.x, the definition of MyClass is slightly different (while MetaClass is not shown because it is unchanged - except that super(MetaClass, cls) can become super() if you want):

#Python 3.x

class MyClass(metaclass=MetaClass):
    pass

As of Python 3.6 there is also a new __init_subclass__ method (see PEP 487) that can be used instead of a meta class (thanks to @matusko for his answer below):

class ParentClass:
    def __init_subclass__(cls, **kwargs):
        super().__init_subclass__(**kwargs)
        register(cls)

class MyClass(ParentClass):
    pass

[edit: fixed missing cls argument to super().__new__()]

[edit: added Python 3.x example]

[edit: corrected order of args to super(), and improved description of 3.x differences]

[edit: add Python 3.6 __init_subclass__ example]