为什么 Haskell 的`show` 函数在这里给我带来麻烦？

Question

我刚接触 Haskell 几天，并从 Learn You a Haskell 中学习它，这对我很有好处。在尝试 99 个 Haskell 问题之一时，我在将函数加载到 ghci 时遇到了以下错误。

问题要求编写一个函数elementAt k x，它接受一个数字k、一个列表x，并提取列表x的kth元素。

这是我的功能

elementAt :: Int -> [a] -> a
elementAt k x
  | k < 0            = error "You have passed a negative index"
  | null x           = error "Cannot extract from an empty list"
  | (length  x) < k  = error "The array contains fewer than " ++ (show k) ++  "elements"
elementAt 0 (x:_)    = x 
elementAt k (_:xs)   = elementAt (k-1) xs 

将此函数加载到 ghci 时出现错误

   Couldn't match expected type `a' with actual type `[Char]'
     `a' is a rigid type variable bound by
         the type signature for elementAt :: Int -> [a] -> a at fun.hs:77:14
   Relevant bindings include
     x :: [a] (bound at fun.hs:78:13)
     elementAt :: Int -> [a] -> a (bound at fun.hs:78:1)
   In the expression:
     error "The array contains fewer than " ++ (show k) ++ "elements"
   In an equation for `elementAt':
       elementAt k x
         | k < 0 = error "You have passed a negative index"
         | null x = error "Cannot extract from an empty list"
         | (length x) < k
         = error "The array contains fewer than " ++ (show k) ++ "elements"

问题似乎在于我使用 show 函数的方式，但我 不明白为什么。在删除 show 调用该函数似乎编译和 完美运行。

Answer 1

您需要在第 5 行的错误消息周围加上括号。

目前您的实现与此相同：

(error "The array contains fewer than ") ++ show k ++  "elements"

虽然您很可能希望它这样做：

error ("The array contains fewer than " ++ show k ++  "elements")

您也可以像这样使用($) 语法：

error $ "The array contains fewer than " ++ show k ++  "elements"

Answer 2

根据Haskell Report，f x ++ g y解析为(f x) ++ (g y)。在你的情况下，

error "The array contains fewer than " ++ (show k) ++  "elements"

解析为

(error "The array contains fewer than ") ++ (show k) ++  "elements"