haskell - 无法从上下文中推导出 (Eq a) (...)

Question

我是 Haskell 的新手。我写了这段代码:

deleteDuplicates :: [a] -> [a]
deleteDuplicates [] = []
deleteDuplicates (x:xs)
        | x == (head xs)        = x : (deleteDuplicates (tail xs))
        | otherwise             = x : (head xs) : (deleteDuplicates (tail xs))

这个错误是什么意思，为什么会发生？我是不是不小心比较了两种不同的类型？

set2.hs:10:3:
    Could not deduce (Eq a) from the context ()
      arising from a use of `==' at set2.hs:10:3-16
    Possible fix:
      add (Eq a) to the context of
        the type signature for `deleteDuplicates'
    In the expression: x == (head xs)
        In a stmt of a pattern guard for
                 the definition of `deleteDuplicates':
          x == (head xs)
    In the definition of `deleteDuplicates':
        deleteDuplicates (x : xs)
                           | x == (head xs) = x : (deleteDuplicates (tail xs))
                           | otherwise = x : (head xs) : (deleteDuplicates (tail xs))

Answer 1

您的类型签名是错误的:

deleteDuplicates :: [a] -> [a]

也就是说你的函数可以处理任何类型的列表，a ，但事实并非如此!你稍后打电话:

x == (head xs)

所以你必须能够比较你的类型是否相等。这意味着签名必须是:

deleteDuplicates :: Eq a => [a] -> [a]

在这种情况下，最好删除您的显式类型签名，在 GHCi 中加载函数并发现解释器认为它应该具有的类型(通过 :t deleteDuplicates)。

更多错误

此外，您对 head 的使用有一个坏主意。 head是一个偏函数，当 xs == [] 时会失败.我建议你扩展模式匹配:

deleteDuplicates (x1:x2:xs)
    | x1 == x2 = ...
    | otherwise = x1 : deleteDuplicates (x2:xs)

还要注意对 otherwise 的修复案件。您跳过了 x2，但是如果 x2 匹配列表中的下一个元素呢？类似 [1,2,2,3]会发现 1 /= 2然后递归会得到列表 [2,3] - 哎呀!