Why are so many built-in types inheritable?
Fabiano Sidler
fabianosidler at my-mail.ch
Sat Mar 18 14:50:48 EST 2006
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.)
Greetings,
F. Sidler
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