haskell - 如何在递归内使用随机数？

我正在尝试从列表中选择一个随机元素，但这会使函数不纯，因此无法编译。如何使递归函数接受IO操作？

build :: Jabberwocky Integer String Syllables -> String
build (Jabberwocky 0 body syl) = body
build (Jabberwocky len body syl)
    | syl == Middle     = build (Jabberwocky (len - 1) (body ++ (rand middle)    ) Consonant)
    | syl == Consonant  = build (Jabberwocky (len - 1) (body ++ (rand consonant)) Vowel)
    | syl == Vowel      = build (Jabberwocky (len - 1) (body ++ (rand vowel)     ) Consonant)
    | syl == Ending     = build (Jabberwocky (len - 1) (body ++ (rand ending)    ) Vowel)
        where
            rand = getStdRandom (randomR (1,6))

最佳答案

您必须将生成器带入纯流程(链接新的随机生成器状态)

randomR_nTimes_rec :: (RandomGen g, Random a) => Int -> (a, a) -> g -> ([a], g)
randomR_nTimes_rec 0 _ g = ([], g)
randomR_nTimes_rec n i g = (x:xs, g'') where ( x, g' ) = randomR i g
                                             (xs, g'') = randomR_nTimes_rec (n - 1) i g'

用法

*Main> getStdGen  >>= return . randomR_nTimes_rec 5 (0,5)
([2,5,3,1,3],1206240749 652912057)

如果您应该将随机状态携带到复杂的过程中，可能会有用Control.Monad.Random with example