Functor instance of generic functors

Question

data Id a = Id a
data Const a b = Const a

the functor instance of above is

instance Functor Id where
  fmap f (Id x ) = Id (f x )
instance Functor (Const a) where
  fmap f (Const x ) = Const x

f somehow did not apply in the const , const (fx) I am confusing about how it work because at least it involves one variable.

Answer 1

Suppose we have f :: Bool -> Int and

x :: Const String Bool
x = Const "some string here"

Now, fmap fx must have type Const String Int , and the only reasonable choice for its result y is

y :: Const String Int
y = Const "some string here"

Note how x and y are roughly the same value, but they belong to distinct types. Further, in order to compute y we do not have to use f in any way, since x does not have any Bool inside, and y does not have any Int inside. f is then irrelevant.

Note that type Const ab is isomorphic to a , whatever type b is. That's why it is named Const : it does not really use its second argument. Its second argument still matters since it causes Const ab and Const a b' to be distinct types, even if both are isomorphic to a .

Answer 2

Const instance just ignores the f function and doesn't change its value. Id instance applies f function to its value and returns new Id with new value.