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我正在尝试编写一个素数生成器并利用 MillerRabin 公式检查数字是否为素数，然后再将数字返回给我。下面是我的代码：

primegen :: Int -> IO Integer
primegen bits =
    fix $ \again -> do
        x <- fmap (.|. 1) $ randomRIO (2^(bits - 1), 2^bits - 1)
        if primecheck x then return x else again

primesTo100 = [2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97]

powerMod :: (Integral a, Integral b) => a -> a -> b -> a
powerMod m _ 0 =  1
powerMod m x n | n > 0 = join (flip f (n - 1)) x `rem` m where
  f _ 0 y = y
  f a d y = g a d where
    g b i | even i    = g (b*b `rem` m) (i `quot` 2)
          | otherwise = f b (i-1) (b*y `rem` m)

witns :: (Num a, Ord a, Random a) => Int -> a -> IO [a]
witns x y = do
     g <- newStdGen 
     let r = [9080191, 4759123141, 2152302898747, 3474749600383, 341550071728321]
         fs = [[31,73],[2,7,61],[2,3,5,7,11],[2,3,5,7,11,13],[2,3,5,7,11,13,17]]
     if  y >= 341550071728321
      then return $ take x $ randomRs (2,y-1) g
       else return $ snd.head.dropWhile ((<= y).fst) $ zip r fs

primecheck :: Integer -> IO Bool
primecheck n | n `elem` primesTo100 = return True
                     | otherwise = do
    let pn = pred n
        e = uncurry (++) . second(take 1) . span even . iterate (`div` 2) $ pn
        try = return . all (\a -> let c = map (powerMod n a) e in
                                  pn `elem` c || last c == 1)
    witns 100 n >>= try

我不明白 IO Bool 发生了什么。我收到以下错误...

 Couldn't match expected type `Bool' with actual type `IO Bool'
 In the return type of a call of `primecheck'
 In the expression: primecheck x
 In a stmt of a 'do' block: if primecheck x then return x else again

如果我将 IO Bool 更改为普通的 Bool，他们会给我这个：

Couldn't match expected type `Bool' with actual type `m0 a0'

谢谢你们的帮助！我很感激。

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2 回答 2

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if primecheck x then return x else again

无效，因为primecheck x返回 type 的值IO Bool。你想用 do 表示法或类似的东西对 monad 进行排序：

primecheck x >>= (\val -> if val then return x else again)
于 2012-11-12T03:23:31.530 回答
3

由于primecheckreturn IO Bool，当你调用它时primegen，你需要对它进行排序而不是像纯函数那样调用它。

primegen :: Int -> IO Integer
primegen bits =
    fix $ \again -> do
        x <- fmap (.|. 1) $ randomRIO (2^(bits - 1), 2^bits - 1)
        success <- primecheck x
        if success then return x else again
于 2012-11-12T03:22:57.507 回答