Applicative functor evaluation is not clear to me

Question

I am currently reading Learn You a Haskell for Great Good! and am stumbling on the explanation for the evaluation of a certain code block. I\'ve read the explanations severa

Answer 1

Let us first take a look how fmap and (<*>) are defined for a function:

instance Functor ((->) r) where
    fmap = (.)

instance Applicative ((->) a) where
    pure = const
    (<*>) f g x = f x (g x)
    liftA2 q f g x = q (f x) (g x)

The expression we aim to evaluate is:

 (+) <$> (+3) <*> (*100)  $ 5

or more verbose:

((+) <$> (+3)) <*> (*100) $ 5

If we thus evaluate (<$>), which is an infix synonym for fmap, we thus see that this is equal to:

(+) . (+3)

so that means our expression is equivalent to:

((+) . (+3)) <*> (*100) $ 5

Next we can apply the sequential application. Here f is thus equal to (+) . (+3) and g is (*100). This thus means that we construct a function that looks like:

\x -> ((+) . (+3)) x ((*100) x)

We can now simplify this and rewrite this into:

\x -> ((+) (x+3)) ((*100) x)

and then rewrite it to:

\x -> (+) (x+3) ((*100) x)

We thus have constructed a function that looks like:

\x -> (x+3) + 100 * x

or simpler:

\x -> 101 * x + 3

If we then calculate:

(\x -> 101*x + 3) 5

then we of course obtain:

101 * 5 + 3

and thus:

505 + 3

which is the expected:

508