What exactly does “deriving Functor” do?

Question

I\'m trying to figure out what exactly are the rules for deriving Functor in Haskell.

I\'ve seen message postings about it, and I\'ve seen test code abo

Answer 1

To use deriving Functor you must enable the DeriveFunctor language pragma and apply it to a polymorphic type which has a covariant final type variable---in other words, a type which admits a valid Functor instance. It'll then derive the "obvious" Functor instance.

There's been some concern in the past that the derived instance is not as efficient as a hand coded one is, though I cannot seem to find that material.

The algorithm itself was, as far as I could find, first proposed by Twan Van Laarhoven in 2007 and makes heavy use of Generic Haskell programming.

Answer 2

The code that actually does the deed is, unfortunately, a bit on the hairy side. I believe that's largely because earlier, simpler code would sometimes lead to excessive compilation times. Twan van Laarhoven came up with the current code to address this.

The derived Functor instance always does the obvious thing. This is usually just fine, but occasionally misses opportunities. For example, suppose I write

data Pair a = Pair a a deriving Functor
data Digit a = One a | Two a a deriving Functor
data Queue a =
    Empty
  | Single a
  | Deep !(Digit a) (Queue (Pair a)) !(Digit a) deriving Functor

This will generate (in GHC 8.2)

instance Functor Queue where
  fmap ...
  x <$ Empty = Empty
  x <$ Single y = Single x
  x <$ Deep pr m sf = Deep (x <$ pr) (fmap (x <$) m) (x <$ sf)

It's possible to write that last case much better by hand:

  x <$ Deep pr m sf = Deep (x <$ pr) (Pair x x <$ m) (x <$ sf)

You can see the actual derived code using -ddump-deriv.