[Haskell-beginners] Monads
Daniel Fischer
daniel.is.fischer at googlemail.com
Wed Mar 2 15:22:35 CET 2011
On Wednesday 02 March 2011 15:03:00, Patrick Lynch wrote:
> In Learn You a Haskell for Great Good!, the author Miran Lipovaca
> indicates that to understand Monads you need to know Functors... So, I
> noticed the following:
>
> class Functor f where
> fmap::(a->b)->f a->f b
>
> instance Functor [] where
> fmap = map
>
> map::(a->b)->[a]->[b]
>
> if [] is substituted for f in the class definition of Functor the
> following is obtained class Functor [] where
> fmap::(a->b)->[]a->[]b
>
> my questions are:
> 1. is this substitution ok?
> 2. is []a = [a]?
> 3. is []b = [b]?
>
> if 2. and 3. are valid then the following is obtained:
> fmap::(a->b)->[a]->[b]
> which is the same as map type and therefore: fmap = map. QED.
>
> Can you please answer questions 2 and 3 above?
>
> Thank you
Yes, that's correct. The list constructor is special ([] isn't valid
Haskell for a type constructor, so it's wired in), it can be used prefix
([] a) or `roundfix' ([a]). Ordinarily, you can use type constructors
prefix or infix (if they take two type arguments), e.g. a -> b = (->) a b;
Either a b = a `Either` b.
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