收到错误 "Non type-variable argument in the constraint: Integral [a2]"

Question

我正在尝试使用以下代码实现 luhn 算法：

luhn :: Int -> Bool
luhn x = (tail $ show (foldl (\acc x -> acc + (read x :: Int)) 0 (foldr doEncrypt [] $ zip [0..] (show x)))) == "0"
    where
        doEncrypt (i,y) acc = if not(even i)
            then head(((uncurry (+) . (`divMod` 10) . (*2)) y)) : acc
            else (head y) : acc

我现在遇到了以下错误：

• Non type-variable argument in the constraint: Integral [a2]
  (Use FlexibleContexts to permit this)
• When checking the inferred type
    doEncrypt :: forall a1 a2.
                 (Integral a1, Integral [a2]) =>
                 (a1, [a2]) -> [a2] -> [a2]
  In an equation for ‘luhn’:
      luhn x
        = (tail
             $ show
                 (foldl
                    (\ acc x -> acc + (read x :: Int))
                    0
                    (foldr doEncrypt [] $ zip [0 .. ] (show x))))
            == "0"
        where
            doEncrypt (i, y) acc
              = if not (even i) then
                    head (((uncurry (+) . (`divMod` 10) . (* 2)) y)) : acc
                else
                    (head y) : acc

我看到错误表明元组的第二部分 (a2) 是 "Non type-variable argument"。但是，Haskell 似乎将此 a2 参数识别为 Integral，而实际上它是 Char。我如何告诉 Haskell 这是一个 Char 而 Haskell 不应该再担心这个变量的类型？还是有其他我不明白的原因导致了这个错误？

编辑：当我删除 (head y) 并将其替换为 y 时，我得到以下错误：

• Couldn't match type ‘Char’ with ‘[Char]’
  Expected type: [String]
    Actual type: [Char]
• In the third argument of ‘foldl’, namely
    ‘(foldr doEncrypt [] $ zip [0 .. ] (show x))’
  In the first argument of ‘show’, namely
    ‘(foldl
        (\ acc x -> acc + (read x :: Int))
        0
        (foldr doEncrypt [] $ zip [0 .. ] (show x)))’
  In the second argument of ‘($)’, namely
    ‘show
       (foldl
          (\ acc x -> acc + (read x :: Int))
          0
          (foldr doEncrypt [] $ zip [0 .. ] (show x)))’

Answer 1

我的解决方案存在多个问题，但最终以下代码有效！

luhn :: Int -> Bool
luhn x = (tail $ show (foldl (\acc x -> acc + (digitToInt x)) 0 (foldr doEncrypt [] $ zip [0..] (show x)))) == "0"
    where
        doEncrypt (i,y) acc = if not(even i)
            then (head $ show(((uncurry (+) . (`divMod` 10) . (*2)) (digitToInt y)))) : acc
            else y : acc

非常感谢@WillemVanOnsem 的指点，否则我可能无法解决这个问题！

收到错误 "Non type-variable argument in the constraint: Integral [a2]"

Getting the error "Non type-variable argument in the constraint: Integral [a2]"

haskell

fold

accumulator