2010/3/20 boblettoj
Ah yes, that makes sense now, however i have another problem, here is the updated code:
--function used to shuffle cards --list equals random member of array plus the rest of the array --i is randomly generated from range of length equal to that of cards. shuffle :: Int -> [a] -> [a] shuffle i [] = [] shuffle i cards = [(cards!!i) : (shuffle (randomR(0, ((length cards)-2))) (delete (cards!!i) cards))]
and the message: cards.hs:32:11: Couldn't match expected type `Int' against inferred type `g -> (Int, g)' In the first argument of `shuffle', namely `(randomR (0, ((length cards) - 2)))' In the second argument of `(:)', namely `(shuffle (randomR (0, ((length cards) - 2))) (delete (cards !! i) cards))' In the expression: (cards !! i) : (shuffle (randomR (0, ((length cards) - 2))) (delete (cards !! i) cards))
Doesn't RandomR return an Int?
No, see docs http://haskell.org/hoogle/?hoogle=randomR
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