Re: [Haskell-beginners] foldr with short circuit and accumulator

Hi, Thanks for your reply. I've done zip'ing in those kinds of problem, also. I think it'd be better if we can do it without zip'ing, with folding only. With help of KwangYul Seo, I know now that short-circuiting with fold is possible, and I made ssfold version of findAdjacent: -- Spencer Janssen's ssfold, from haskell-cafe ssfold p f a0 xs = foldr (\x xs a -> if p a then a else xs (f a x)) id xs a0 findAdjacent :: [Int] -> Maybe Int findAdjacent = snd . ssfold (uncurry const) step (False, Nothing) where step acc@(True, _) _ = acc step (False, last) x = (if Just x == last then True else False, Just x) findAdjacent $ [1..10000] ++ [10000..] ==> 10000 It might be overkill for finding adjacent entry. It's better to use zip'ing. But I think ssfold can be useful in other cases. Any comments will be welcomed. Regards, Chul-Woong 2016-02-02 17:47 GMT+09:00 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 wrote:

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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.

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