[Haskell-beginners] foldr with short circuit and accumulator

[Chul-Woong Yang]

Thank you for introducing mind-blowing ssfold, which has step function
with three argument:

> ssfold p f a0 xs = foldr (\x xs a -> if p a then a else xs (f a x)) id xs a0

Having spent last night, I've yet to got it. What a slow person!

Regards,

2016-02-02 18:03 GMT+09:00 KwangYul Seo <kwangyul.seo at gmail.com>:
> Hi,
>
> Implementing a short-circuiting fold was already discussed in the
> haskell-cafe mailing list.
>
> https://mail.haskell.org/pipermail/haskell-cafe/2007-April/024171.html
>
>
> On Tue, Feb 2, 2016 at 4:15 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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