[Haskell-cafe] recursion issues...

[Andrew Wagner]

Looks to me like you want:
poolNewsB = foldr poolNews (0,0,0,0)

On Nov 10, 2007 11:54 AM, Ryan Bloor <ryanbloor at hotmail.com> wrote:
>
> hiya
>
>  I was wondering how I would get the second function do recursively do the
> function for poolNews xs.... tried that and it fails.
>
>  Ryan
>
>
>
>
> --Give wins, draws a rating.
>
>
>
> poolNews :: Result -> PoolNews -> PoolNews
>
> poolNews (a,b,c,d,e) (home,away,goaless,scoredraw)
>
>              | c > d = (home+1,away,goaless,scoredraw)
>
>              | c < d = (home,away+1,goaless,scoredraw)
>
>              |(c == 0) && (d == 0) = (home+1,away,goaless+1,scoredraw)
>
>                   | otherwise = (home,away,goaless,scoredraw+1)
>
>
>
>
>
> --Do for all Results
>
> poolNewsB :: Results -> PoolNews poolNewsB (x:xs) = poolNews x (0,0,0,0)
>
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