Python "implements <interface>" equivalent?
pavlovevidence at gmail.com
Thu Oct 4 17:27:15 CEST 2007
On Oct 4, 11:11 am, Wojciech Gryc <wojci... at gmail.com> wrote:
> I recently started using Python and am extremely happy with how
> productive it's made me, even as a new user. I'm hoping to continue
> using the language for my research, and have come across a bit of a
> stumbling block.
> I'm a seasoned Java programmer and quite a big fan of interfaces...
> i.e. The idea that if I make a number of distinct classes that
> implement interface X, I can pass them all as parameters to functions
> or whatnot that require an X object.
> Is there something similar in Python?
Yes: you do it pretty much the same way you'd do it in Java, except
for two differences:
* leave out the "implements Interface" part
* don't actually create an Interface class
And then, voila!, you can write functions that can accept any class
that implements your interface.
> What I'd like to do is create a feature detection system for my work
> -- specifically, a general class / interface called "Feature" and then
> subclasses that implement functions like isFeaturePresent() in all of
> their different and unique ways. I'd love to hear how I can do this in
Just define isFeaturePresent() in any class you want. That's all you
have to do; any class that defines this method can be passed to any
function that invokes it, without having to "implement" or "subclass"
This is known as "duck typing" in Python lingo.
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