Question

这是原始问题：

修改问题10的结果，使得如果元素没有重复，则只需将其复制到结果列表中。只有带有重复项的元素才会作为（N E）列表传输。

示例：

* (encode-modified '(a a a a b c c a a d e e e e))
((4 A) B (2 C) (2 A) D (4 E))

Haskell中的示例：

ghci> encodeModified "aaaabccaadeeee"
[Multiple 4 'a',Single 'b',Multiple 2 'c',
Multiple 2 'a',Single 'd',Multiple 4 'e']

这是我的解决方案：

import Data.List

data Encode a = Single a | Multiple Int a deriving (Show)

encodeModified :: (Eq a) => [a] -> [Encode a]
encodeModified xs = [y | x <- group xs, let y = if (length x) == 1   then Single (head x) else Multiple (length x) (head x)]

encodeModified' = map (\xs -> if (length xs == 1) then Single (head xs) else Multiple (length xs) (head xs)) . group

我的GHCI是7.10.3，当我执行：l xx.hs时，它给了我：

No instance for (Eq a0) arising from a use of ‘group’
    The type variable ‘a0’ is ambiguous
    Relevant bindings include
      encodeModified' :: [a0] -> [Encode a0] (bound at 11.hs:8:1)
    Note: there are several potential instances:
      instance (Eq a, Eq b) => Eq (Either a b)
        -- Defined in ‘Data.Either’
      instance forall (k :: BOX) (s :: k). Eq (Data.Proxy.Proxy s)
        -- Defined in ‘Data.Proxy’
      instance (GHC.Arr.Ix i, Eq e) => Eq (GHC.Arr.Array i e)
        -- Defined in ‘GHC.Arr’
      ...plus 27 others
    In the second argument of ‘(.)’, namely ‘group’
    In the expression:
      map
        (\ xs
           -> if (length xs == 1) then
                  Single (head xs)
              else
                  Multiple (length xs) (head xs))
      . group
    In an equation for ‘encodeModified'’:
        encodeModified'
          = map
              (\ xs
                 -> if (length xs == 1) then
                        Single (head xs)
                    else
                        Multiple (length xs) (head xs))
            . group

我无法通过反馈找出错误是什么，如果有人可以帮助我，我的方法没有发现任何问题？感谢。

Answer 1

此处的代码包含Eq a类型的encodeModified约束，但是根据您发布的错误，我可以看到您文件中实际存在encodeModified'的类型（注意素数！）没有那个约束。添加约束，你应该很高兴。

将来，发布实际有问题的代码是礼貌的。

Haskell中的类型变量不明确

1 个答案: