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

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 :

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 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

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