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