Fabiano Sidler wrote: > Hi folks! > > For debugging purposes I tried this: > > --- snip --- > def foo(): pass > function = type(foo) > > class PrintingFunction(function): > def __init__(self, func): > self.func = func > def __call__(self, *args, **kwargs): > print args, kwargs > return function.__call__(self, args, kwargs) > > class DebugMeta(type): > def __new__(self, name, bases, dict): > for name in dict: > if type(dict[name]) is function: > dict[name] = PrintingFunction(dict[name]) > > --- snap --- > > Now I tought I were able to let all maethod of classes with DebugMeta as > metaclass print out their arguments. But I got the following sad error: > > TypeError: Error when calling the metaclass bases > type 'function' is not an acceptable base type > > That's awful, isn't it? > What could I do to get the above code working? (No, I disliked to re- > implement <type 'function'> without this unpleasant behaviour in Python.)
You could do this with a simple decorator: http://wiki.python.org/moin/PythonDecoratorLibrary#head-d4ce77c6d6e75aad25baf982f6fec0ff4b3653f4 or I think your class PrintingFunction would work as class PrintingFunction(object): def __init__(self, func): self.func = func def __call__(self, *args, **kwargs): print args, kwargs return self.func(*args, **kwargs) Kent -- http://mail.python.org/mailman/listinfo/python-list
