Carl Banks a écrit :
(snip)
class A(object):
def __init__(self,*args,**kwargs):
raise TypeError('Type not callable; use factory function
instead')
@classmethod
def _create_object(cls,initial_value):
self = object.__new__(cls) # avoid __init__
self.value = in
On Mon, 10 Aug 2009 16:37:25 +0200, Ulrich Eckhardt wrote:
> ...that is the question!
>
> I have a module which exports a type. It also exports a function that
> returns instances of that type. Now, the reason for my question is that
> while users will directly use instances of the type, they wil
On Aug 10, 7:37 am, Ulrich Eckhardt wrote:
> ...that is the question!
>
> I have a module which exports a type. It also exports a function that
> returns instances of that type. Now, the reason for my question is that
> while users will directly use instances of the type, they will not create
> in
10-08-2009 Ulrich Eckhardt wrote:
So, the type is a part of the public API, but its constructor is not.
Should
I mark the type as private (with a leading underscore) or not?
IMHO you shouldn't (i.e. name should be marked "public") because of
possible usage of e.g. "isinstance(foo, YourType)
