Checking for required arguments when instantiating class.

[Lacrima]

On May 6, 3:36 pm, Chris Rebert <c... at rebertia.com> wrote:
> On Wed, May 6, 2009 at 5:24 AM, Piet van Oostrum <p... at cs.uu.nl> wrote:
>
>
>
> >>>>>> Lacrima <Lacrima.Ma... at gmail.com> (L) wrote:
>
> >>L> Hello!
> >>L> For example I have two classes:
>
> >>>>>> class First:
> >>L>     def __init__(self, *args, **kwargs):
> >>L>             pass
>
> >>>>>> class Second:
> >>L>     def __init__(self, somearg, *args, **kwargs):
> >>L>             self.somearg = somearg
>
> >>L> How can I test that First class takes 1 required argument and Second
> >>L> class takes no required arguments?
> >>L> So that I could instantiate them in a for loop.
>
> >>>>>> a = [First, Second]
> >>>>>> for cls in a:
> >>L>     instance = cls()
>
> >>L> Traceback (most recent call last):
> >>L>   File "<pyshell#22>", line 2, in <module>
> >>L>     instance = cls()
> >>L> TypeError: __init__() takes at least 2 arguments (1 given)
>
> >>L> Of course, I can do like this:
> >>>>>> for cls in a:
> >>L>     try:
> >>L>             instance = cls()
> >>L>     except TypeError:
> >>L>             instance = cls('hello')
>
> >>>>>> print instance.somearg
> >>L> hello
>
> >>L> But what if I have to instantiate any class with 3 or 4 required
> >>L> arguments? How can I do it?
>
> > cls.__init__.im_func.__code__.co_argcount
>
> > This will include self, so it will be 1 in First and 2 in Second.
>
> AFAICT, that would count non-required arguments too, which isn't
> strictly what the OP requested.
>
> > However this is very dirty trickery and should not be recommended. It
> > may also change in future versions and other implementations of Python.
>
> Very much agreed.
>
> > I think it would be cleaner to put a class attribute in the classes that
> > defines how they should be initialized (e.g. just the number of required
> > arguments or more specific information) or have a special factory method
> > for this use case.
>
> Seconded. I'd recommend the latter personally, though it's impossible
> to give a definitive answer without more context.
>
> Cheers,
> Chris
> --http://blog.rebertia.com

Thanks for all of you!

I think I'll try to write a special method for this case and will
report you result.

-Max