If you mean is there any f and z for which this can be done with only "foldr f z xs", I believe the answer is no. If you don't mind extra parts, though: findAdjacent :: (Eq a) => [a] -> Maybe a findAdjacent xs = foldr f Nothing $ zip xs ps where ps = zipWith (==) (tail xs) xs f (x,p) next = if p then Just x else next On Mon, Feb 1, 2016 at 11:15 PM, Chul-Woong Yang <cwyang@aranetworks.com> wrote:
Hi, all.
Can it be possible to do fold with short circuit and accumulator both? For example, can we find an element which is same value to adjacent one?
findAdjacent [1,2..n, n, n+1, n+2.......] => n \__very long__/
Though there can be many ways to do it, Can we do it with fold[r|l]?
I'll be happy to receive any comments.
Chul-Woong _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners