I'm reading the "Haskell programming" book and in chapter 16 about functors they combine two "fmap" with the "(.)" function.
I really don't understand how the type of "fmap . fmap" is "(Functor f2, Functor f1) => (a -> b) -> f1 (f2 a) -> f1 (f2 b)". The thing that really freaks me out is that I thought that "(.)" arguments are ONLY two unary function. Maybe there is some curring magic underline, there is someone that can explain to me how the type of "fmap . fmap" is derived from the type of "fmap" and "(.)"? _______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
Hi, as you probably know, fmap ha type fmap :: Functor f => (a -> b) -> f a -> f b So, you can see fmap as a unary funcrion that takes an argument of type (a->b) and returns a result of type (f a -> f b). For example, if you define identity x = x then fmap identity :: Functor f => f b -> f b (here a = b because identity :: p -> p ) Now, the type of (.) is; (.) :: (b -> c) -> (a -> b) -> (a -> c) if you write
fmap . fmap, in general (a priori) the type of the fmaps could be FIRSTFMAP fmap :: Functor f1 => (d -> e) -> f1 d -> f1 e SECONDFMAP fmap :: Functor f2 => (g -> h) -> f2 g -> f2 h let's see if there must be some bonds between types d,e,g,h, due to the type of (.) (b -> c) = FIRSTFMAP (the first argument of (.) ) (a -> b) = SECONDFMAP
(the second argument of (.) )
so, we have (I'll not write "Functor f1" or "Functor f2", we know f1 and f2 are functors) b = (d -> e) c = (f1 d -> f1 e) a = (g -> h) b = (f2 g -> f2 h) now, it must be b=b, so the type (d -> e) must be the same type of (f2 g -> f2 h) So d = f2 g e = f2 h The result of fmap . fmap, will be of type (a -> c) that is (g -> h) -> (f1 d -> f1 e) = (g -> h) -> f1 d -> f1 e that, remebering the conditions on d and e, is (g -> h) -> f1 (f2 g) -> f1 (f2 h) which is the type you found on the book (you only need to add that f1, f2 are Functors, and the names of f and g can of course be replaced with a and b) Cheers, Ut 2018-06-12 18:12 GMT+02:00 Marco Turchetto <[hidden email]>:
_______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
It's all clear as day. Thanks On Tue, 12 Jun 2018 at 19:35 Ut Primum <[hidden email]> wrote:
_______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners |
Free forum by Nabble | Edit this page |