[Haskell-beginners] fmap composition

[Marco Turchetto]

It's all clear as day.

Thanks

On Tue, 12 Jun 2018 at 19:35 Ut Primum <utprimum at gmail.com> wrote:

> 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 <marco.turchetto.mt at gmail.com>:
>
>> 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
>> Beginners at haskell.org
>> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>>
>>
> _______________________________________________
> Beginners mailing list
> Beginners at haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://mail.haskell.org/pipermail/beginners/attachments/20180613/d160adaf/attachment-0001.html>