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by Y_Y 630 days 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.
1 comments
Iceland_jack 628 days 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). 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 ($))
    = ($)
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Iceland_jack 628 days 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-...

This is valid definition of lift along a different, less interesting but more general axis.

    lift = (??) @[]
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