[Haskell-beginners] Functions as Applicatives

[Olumide]

I must be missing something. I thought f accepts just one argument.

- Olumide

On 23/08/2016 00:54, Theodore Lief Gannon wrote:
> Yes, (g x) is the second argument to f. Consider the type signature:
>
> (<*>) :: Applicative f => f (a -> b) -> f a -> f b
>
> In this case, the type of f is ((->) r). Specialized to that type:
>
> (<*>) :: (r -> a -> b) -> (r -> a) -> (r -> b)
> f <*> g = \x -> f x (g x)
>
> Breaking down the pieces...
> f :: r -> a -> b
> g :: r -> a
> x :: r
> (g x) :: a
> (f x (g x)) :: b
>
> The example is made a bit confusing by tossing in an fmap. As far as the
> definition above is concerned, 'f' in the example is ((+) <$> (+3)) and
> that has to be resolved before looking at <*>.
>
>
> On Mon, Aug 22, 2016 at 9:07 AM, Olumide <50295 at web.de
> <mailto:50295 at web.de>> wrote:
>
>     Hi List,
>
>     I'm struggling to relate the definition of a function as a function
>
>     instance Applicative ((->) r) where
>         pure x = (\_ -> x)
>         f <*> g = \x -> f x (g x)
>
>     with the following expression
>
>     ghci> :t (+) <$> (+3) <*> (*100)
>     (+) <$> (+3) <*> (*100) :: (Num a) => a -> a
>     ghci> (+) <$> (+3) <*> (*100) $ 5
>     508
>
>     From chapter 11 of LYH http://goo.gl/7kl2TM .
>
>     I understand the explanation in the book: "we're making a function
>     that will use + on the results of (+3) and (*100) and return that.
>     To demonstrate on a real example, when we did (+) <$> (+3) <*>
>     (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in
>     8 and 500. Then, + gets called with 8 and 500, resulting in 508."
>
>     The problem is that I can't relate that explanation with the
>     definition of a function as an applicative; especially f <*> g = \x
>     -> f x (g x) . Is (g x) the second argument to f?
>
>     Regards,
>
>     - Olumide
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