module RandomProgression where import Random data RandomProgression r a = RP (r -> (a, r)) nextRandom :: (RandomGen r, Random a) => RandomProgression r a nextRandom = RP (\r -> random r) instance Monad RandomProgression r a where --(>>=) :: RandomProgression r a -> (a -> RandomProgression r b) -> RandomProgression r b RP x >>= f = RP (\r -> let (a, r') = x r in let RP b = f a in b r') --return :: a -> RandomProgression r a return value = RP (\r -> (value, r))
書いてみました…instanceでエラーが出ます…わかりません…
↓こんな風に使えたらいいなと思ってる let (answer, g) = (do a <- nextRandom; return (a `mod` 10)) (mkStdGen 100)
736 名前：デフォルトの名無しさん mailto:sage [03/07/21 02:45]
括弧が要るんですね… instance Monad (RandomProgression r) where 実行用に eval :: (RandomGen r) => RandomProgression r a -> r -> (a, r) eval (RP e) r = e r 書き足して RandomProgression> eval((do x <- nextRandom; return (x `mod` 10)) :: RandomProgression StdGen Int) (mkStdGen 10) (7,432453652 1655838864) 動いたー！
>>734 知らん。気になるのなら、たとえば、こうすればちゃんとモナドになる。 module Main where instance (Num a) => Monad ((,) a) where (>>=) (sub, main) f = (sub+sub', main') where (sub', main') = f main return main = (0, main)
>>733 まずちょっと修正...ｽﾏｿ。 import Monad -- 追加 instance Monad ReadR where ReadR r >>= fr = ReadR (\str -> [ret| (v,str') <- r str, ret <- fr v `runr` str']) return v = ReadR (\str -> [(v, str)]) fail s = mzero -- 以下追加 instance MonadPlus ReadR where mzero = ReadR (\s -> []) mplus (ReadR f) (ReadR g) = ReadR (\s -> f s ++ g s)
上のコードのReadR版 >rreadTree :: (Read a) => ReadR (Tree a) >readsTree s = rreadTree `runr` s >rreadTree = do "[" <- rlex > l <- rreadTree > "|" <- rlex > r <- rreadTree > "]" <- rlex > return $ Branch l r > `mplus` > do x <- rread > return $ Leaf x
