confused about function as instance of Functor in haskell

Question

the type of fmap in Functor is:

fmap :: Functor f => (a -> b) -> f a -> f b

it looks like ,first apply function (a -> b) to the

Answer 1

The fmap instance for (->) r (i.e. functions) is literally just composition. From the source itself:

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

So, in your example, we can just replace fmap with (.), and do some transformations

fmap (*3) (+100) 1 => 
(.) (*3) (+100) 1  =>
(*3) . (+100) $ 1  => -- put (.) infix
(*3) (1 + 100)     => -- apply (+100)
(1 + 100) * 3         -- apply (*3)

That is, fmap for functions composes them right to left (exactly the same as (.), which is sensible because it is (.)).

To look at it another way (for (double) confirmation!), we can use the type signature:

-- general fmap
fmap :: Functor f => (a -> b) -> f a -> f b

-- specialised to the function functor (I've removed the last pair of brackets)
fmap :: (a -> b) -> (r -> a) -> r -> b 

So first the value of type r (the third argument) needs to be transformed into a value of type a (by the r -> a function), so that the a -> b function can transform it into a value of type b (the result).