On Nov 10, 2007 11:54 AM, Ryan Bloor <[EMAIL PROTECTED]> 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) >
As Andrew points out, really what you want is a fold. However, perhaps you're not supposed to use such Prelude functions in your assignment? The real issue here is that your definition of poolNewsB does not do anything with xs. You need to somehow incorportate a recursive call that will process the rest of the list (xs) as well as the first item (x). Hope that helps! -Brent
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