functions as applicative functors (Haskell / LYAH)

Question

Chapter 11 of Learn You a Haskell introduces the following definition:

instance Applicative ((->) r) where
    pure x = (\\_ -> x)
    f <*         


        

Answer 1

Going through your original question, I think there's one subtle but very key point that you might have missed. Using the original example from LYAH:

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

This is the same as:

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

The key here is the pure before (+), which has the effect of boxing (+) as an Applicative. If you look at how pure is defined, you can see that to unbox it, you need to provide an additional argument, which can be anything. Applying <*> to (+) <$> (+3), we get

\x -> (pure (+)) x ((+3) x)

Notice in (pure (+)) x, we are applying x to pure to unbox (+). So we now have

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

Adding (*100) to get (+) <$> (+3) <*> (*100) and apply <*> again, we get

\y -> (\x -> (+) ((+3) x)) y ((*100) y) {Since f <*> g = f x (g x)}

5  -> (\x -> (+) ((+3) x)) 5 ((*100) 5)

(\x -> (+) ((+3) x)) 5 (500)

5 -> (+) ((+3) 5) (500)

(+) 8 500

508

So in conclusion, the x after f is NOT the first argument to our binary operator, it is used to UNBOX the operator inside pure.