Why does Functor not expose a default implementation of fmap?

Question

In the Functor class type definition:

class Functor f where
    fmap :: (a -> b) -> f a -> f b

Why does fmap not have a default implementation? Something like that:

class Functor f where
    fmap :: (a -> b) -> f a -> f b
    fmap fn (f a) = (f $ (fn) a)

When I write instances of Functor, I write the same code for each instance manually. Is there a way for me to specify a default implementation?

Answer 1

You probably want this:

{-# LANGUAGE DeriveFunctor #-}

data T1 a = T1 a
  deriving Functor
data T2 a = T2 a
  deriving Functor

As to why there's no default implementation for functor: your proposal only works if f is the identity functor (up to isomorphism). That is, it works on

data F a = F a

but it is not going to work on

data F a = F a a

or

data F a = F (Int -> a) [a] (Maybe a)

which require more complex fmaps.

While one can not write a default fmap which works in every case, in many simple cases such as the above ones it seems trivial to spot what fmap should be. Fortunately, we now have derive Functor which covers these simple cases.