Haskell 调用预期类型的方程时出错

Question

我正在执行以下对 haskell 方法的调用

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：错误： • 预期类型为“* -> Constraint”，但“Int”具有类型“*” • 在类型签名中： rollD6 :: (RandomGen g, Int a) => Int -> ([a], g) | 11 | 11 rollD6 :: (RandomGen g, Int a) => Int -> ([a], g) |
^^^^^

src/Merp/Die/Boundary/RollADie.hs:14:32：错误： • 预期类型为“* -> Constraint”，但“Int”具有类型“*” • 在类型签名中： rollD6InRange :: (RandomGen g, Int a) => Int -> Int -> Int -> ([a], g) | 14 | 14 rollD6InRange :: (RandomGen g, Int a) => Int -> Int -> Int -> ([a], g) | ^^^^^

我一直在研究调用的参数，但我一直无法弄清楚发生了什么。你能告诉我还剩下什么吗？

Answer 1

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：你用

a

代替

Int

，所以：

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

同样适用于

rollD6InRange

。