[Haskell-beginners] foldr with short circuit and accumulator

[Theodore Lief Gannon]

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 at 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
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