10 ms·

lift = sequenceA @[] @(a ->)

• Y_Y 2y ago
  Did you do that by hand? I guessed you used the classic `pointfree` program, bit that gave lift = fix ((`ap` tail) . (. head) . flip ((.) . liftM2 (:))) As far as I know it's not possible to get this functionality in Haskell even with clever instance magic, but I'd love to be proved wrong.
  • Iceland_jack 2y ago
    > As far as I know it's not possible to get this functionality in Haskell even with clever instance magic It is possible to fill in basic function bodies based on their type, using ghc-justdoit (https://hackage.haskell.org/package/ghc-justdoit https://hackage.haskell.org/package/ghc-justdoit). That's maybe not what you meant, if you are looking for integrating pointfree into Haskell it can be added to ghci or your development environment. foo :: ((a -> r) -> r) -> (a -> ((b -> r) -> r)) -> ((b -> r) -> r) foo = (…) In this case I wrote it because I knew about the pattern. Your lift definition is just ($) flip \a -> ($ a) = flip (&) = flip (flip ($)) = ($)
    • Iceland_jack 2y ago
      Maybe you meant to write (??) :: Functor f => f (a -> b) -> a -> f b funs ?? a = fmap ($ a) funs from lens: https://hackage.haskell.org/package/lens-5.3.2/docs/Control-Lens-Lens.html#v:-63--63- https://hackage.haskell.org/package/lens-5.3.2/docs/Control-... This is valid definition of lift along a different, less interesting but more general axis. lift = (??) @[]
• Y_Y 2y ago
  lift = flip $ \a -> ($a)