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Yago
Asked: 2024-05-24 02:54:25 +0800 CST

Haskell 调用预期类型的​​方程时出错

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我正在对 haskell 方法执行以下调用

rollD6 ::  (RandomGen g, Int a) => Int -> ([a], g)
rollD6 times =  roll Die6 times (mkStdGen 6)

rollD6InRange :: (RandomGen g, Int a) => Int -> Int -> Int -> ([a], g)
rollD6InRange times start end = rollR Die6 times start end (mkStdGen 6)

class Roll a where
  roll :: (RandomGen g) => a -> Int -> g -> ([Int], g)
  rollR :: (RandomGen g) => a -> Int -> Int -> Int -> g -> ([Int], g)

instance Roll Die where
  roll Die6 times = generateIntRandomsArray 1 6 times
  roll Die4 times = generateIntRandomsArray 1 4 times
  roll Die20 times = generateIntRandomsArray 1 20 times
  roll Die100 times = generateIntRandomsArray 1 100 times

  rollR Die6  times first last = generateIntRandomsArray first last times
  rollR Die4 times first last = generateIntRandomsArray first last times
  rollR Die20 times first last = generateIntRandomsArray first last times
  rollR Die100 times first last = generateIntRandomsArray first last times

generateIntRandomsArray :: (Eq a, Random a, Num a, RandomGen g) => a -> a -> a -> g -> ([a], g)
generateIntRandomsArray start end = generate
    where generate 0 generator = ([], generator)
          generate numberElements generator =
              let (value, newGenerator) = randomR (start,end) generator
                  (restOfList, finalGenerator) = generate (numberElements-1) newGenerator
              in  (value:restOfList, finalGenerator)

我收到以下错误

src/Merp/Die/Boundary/RollADie.hs:11:26: error:
    • Expected kind ‘* -> Constraint’, but ‘Int’ has kind ‘*’
    • In the type signature:
        rollD6 :: (RandomGen g, Int a) => Int -> ([a], g)
   |
11 | rollD6 ::  (RandomGen g, Int a) => Int -> ([a], g)
   |                          ^^^^^

src/Merp/Die/Boundary/RollADie.hs:14:32: error:
    • Expected kind ‘* -> Constraint’, but ‘Int’ has kind ‘*’
    • In the type signature:
        rollD6InRange :: (RandomGen g, Int a) =>
                         Int -> Int -> Int -> ([a], g)
    |
 14 | rollD6InRange :: (RandomGen g, Int a) => Int -> Int -> Int -> ([a], g)
    |                                ^^^^^

我一直在研究调用的参数，但我无法弄清楚发生了什么。你能告诉我还剩下什么吗？

haskell

1 个回答

  1. Best Answer

    Int不是类型类，所以：

    rollD6 :: (RandomGen g, Int a) => Int -> ([a], g)
rollD6 times = roll Die6 times (mkStdGen 6)

    没有多大意义：你替换a，Int所以：

    rollD6 :: RandomGen g => Int -> ([Int], g)
rollD6 times = roll Die6 times (mkStdGen 6)

    这同样适用于rollD6InRange.

    我还建议将范围类型和元素数量分开，因此：

    generateIntRandomsArray :: (Random a, Integral b, RandomGen g) => a -> a -> b -> g -> ([a], g)
generateIntRandomsArray start end = generate
  where
    generate 0 generator = ([], generator)
    generate numberElements generator =
      let (value, newGenerator) = randomR (start, end) generator
          (restOfList, finalGenerator) = generate (numberElements -1) newGenerator
       in (value : restOfList, finalGenerator)

    Char例如，这允许我们从s、Bools、Floats 或任何其他[Hackage]实例的序列生成，例如：Random 

    ghci> generateIntRandomsArray 'a' 'z' 10 (mkStdGen 0)
("nlrkwldbji",1695805043 1336516156)
    • 3

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