Re: [Haskell-cafe] Questions about the Functor class and it's use in "Data types à la carte"
Corey O'Connor
coreyoconnor at gmail.com
Mon Dec 17 14:40:17 EST 2007
Excellent! This has all been very helpful. Thanks a lot everybody! :-)
-Corey
On 12/14/07, Benja Fallenstein <benja.fallenstein at gmail.com> wrote:
> Hi Corey,
>
> On Dec 14, 2007 8:44 PM, Corey O'Connor <coreyoconnor at gmail.com> wrote:
> > The reason I find all this odd is because I'm not sure how the type
> > class Functor relates to the category theory concept of a functor. How
> > does declaring a type constructor to be an instance of the Functor
> > class relate to a functor? Is the type constructor considered a
> > functor?
>
> Recall the definition of functor. From Wikipedia:
>
> "A functor F from C to D is a mapping that
>
> * associates to each object X in C an object F(X) in D,
> * associates to each morphism f:X -> Y in C a morphism F(f):F(X)
> -> F(Y) in D
>
> such that the following two properties hold:
>
> * F(idX) = idF(X) for every object X in C
> * F(g . f) = F(g) . F(f) for all morphisms f:X -> Y and g:Y -> Z."
>
> http://en.wikipedia.org/wiki/Functor
>
> We consider C = D = the category of types. Note that any type
> constructor is a mapping from types to types -- thus it associates to
> each object (type) X an object (type) F(X).
>
> Declaring a type constructor to be an instance of Functor means that
> you have to provide 'fmap :: (a -> b) -> (f a -> f b)" -- that is, a
> mapping that associates to each morphism (function) "fn :: a -> b" a
> morphism "fmap fn :: f a -> f b".
>
> Making sure that the two laws are fulfilled is the responsibility of
> the programmer writing the instance of Functor. (I.e., you're not
> supposed to do this: instance Functor Val where fmap f (Val x) = Val
> (x+1).)
>
> Hope this helps with seeing the correspondence? :-)
> - Benja
>
--
-Corey O'Connor
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