Haskell Foldable Wats

David Feuer david.feuer at gmail.com
Wed Feb 24 16:38:56 UTC 2016

On Wed, Feb 24, 2016 at 11:22 AM, Kosyrev Serge <_deepfire at feelingofgreen.ru
> wrote:

> Are you also saying that this cannot be resolved by some kind of a
> type families-based type-level 'flip'?

It really can't, no. You can define

type family FlipF f a b where
  FlipF f a b = f b a

but FlipF, being a type family, is not first-class. You can't make any
instances whatsoever for FlipF f a -- you'll probably get an error about a
partially applied type family. What you *can* do is make a Flip newtype:

newtype Flip f a b = Flip {unflip :: f b a}

This behaves perfectly:

instance Bifunctor f => Functor (Flip f a) where
  fmap f = Flip . first f . unflip
instance Profunctor f => Contravariant (Flip f a) where
  contramap f = Flip . lmap f . unflip
instance Bifunctor f => Bifunctor (Flip f) where
  bimap f g (Flip x) = Flip (bimap g f x)
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