[Haskell-beginners] Functions as Applicatives

[Theodore Lief Gannon]

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> 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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