[Haskell-beginners] foldr with short circuit and accumulator

[Chul-Woong Yang]

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 <tanuki at gmail.com>:
> 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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>
>
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