Python 3 dynamically assign base class

Question

I am writing a simple chess program to practice my OOP in python 3 and was wondering how to dynamically change (before class creation) the base class for a class definition. My class structure is this.

abstract Piece class -> various derived pieces
Board class, has a composite of derived Pieces, and 8x8 matrix, and some methods
abstract Interface class -> CLI or
abstract Interface class -> GUI (also subclassing Tkinter)
Game class (for processing the game logic and main loop), which currently has a Board class member.

I initially implemented the Game class as having an interface data member that is defined during init but I'm finding myself sending a lot of the other internal Game data to the Interface composite member. I feel it would be more elegant to have the Game class be a subclass of either Interface subclass so the it could access their methods directly (and make them abstract).

However I want a version of the Game class that can do this dynamically so that I don't have to code it twice or inherit from both and make runtime decisions on which base class to use. I've currently done this by nesting the Game class inside a function like so.

def Game(ui):
    class Game(ui):
        ...
    return Game()

The crummy naming is part of the reason I don't like this solution. I want to be able to call the Game class on its own without explicitly using or acknowledging that I'm doing anything out of the ordinary.

Is there a way to do this with a metaclass or a class decorator? I have only been able to get them to affect class attributes, not the parent classes.

John La Rooy · Accepted Answer · 2013-05-21 00:12:03Z

4

The class statement is "syntactic sugar" for

type(name, bases, dict)

You can create such a dynamic class using type like this

>>> class ui():
...     def start(self): print("Started!")
... 
>>> Game = type("Game", (ui,), {})
>>> game = Game()
>>> game.start()
Started!

answered May 21, 2013 at 0:12

John La Rooy

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Kevin Over a year ago

Thanks @gnibbler, but I had to modify this a little to keep any of the new attributes that the Game class introduced by using Game = type("Game", (ui,), dict(Game.__dict__)) instead. Also, I hadn't realized the syntactic sugar of the class statement though, so thanks for that!

mgilson · Accepted Answer · 2013-05-21 00:12:04Z

3

You could use a very simple metclass here, but that's overkill. You can just swap out which class you're using as the base class based on whatever condition you want:

>>> class Foo: pass
... 
>>> class Bar: pass
... 
>>> x = 3
>>> class Game(Foo if x < 3 else Bar):pass
... 
>>> Game.__bases__
(<class '__main__.Bar'>,)

Note that this isn't really any different than the formalism you have. However, if I was to use your code, I wouldn't create the Game class and the instance all in the function. I would do something like:

def Game_Factory(base):
    class Game(base):
         ...
    return Game

Game1 = Game_Factory(base1)
Game2 = Game_Factory(base2)

game1_instance = Game1()
game2_instance = Game2()

This gives you much easier access to the Game class (rather than needing to inspect an instance to get it).

answered May 21, 2013 at 0:12

mgilson

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1 Comment

Kevin Over a year ago

This is probably the cleanest way of doing this outside of refactoring my class setup.

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