Oops, I thought I had sent a response to the cafe, but it looks like it just went to Matthew.
Unfortunately, I was trying to give a simplification of the real problem, where the monad is STM instead of []. Based on apfelmus's observation of why they can't be isomorphic, I'm guessing I'm out of luck. http://www.haskell.org/pipermail/haskell-cafe/2006-December/020041.html So in reality, I'm trying to construct something like f :: (a -> STM b) -> STM (a -> b) I just figured it was a general monadic kind of problem, more simply expressed using lists. But the (!!) solution doesn't make sense in this context. On 2/2/07, Yitzchak Gale <[EMAIL PROTECTED]> wrote:
Chad Scherrer wrote: > Are (a -> [b]) and [a -> b] isomorphic? I'm trying to construct a function > > f :: (a -> [b]) -> [a -> b] > > that is the (at least one-sided) inverse of > > f' :: [a -> b] -> a -> [b] > f' gs x = map ($ x) gs Anything better than this? f g = [\x -> g x !! n | n <- [0..]] -Yitz
-- Chad Scherrer "Time flies like an arrow; fruit flies like a banana" -- Groucho Marx _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
