此代码中预期的 Haskell 反向函数是什么？

Question

我正在写一个小的Haskell练习，它应该移动列表中的一些元素，类似于凯撒密码，代码已经可以运行，代码如下。

module Lib (shift, cipherEncode, cipherDecode ) where
import Data.Char
import Data.List
import Data.Maybe

abcdata :: [Char]
abcdata = ['a','b','c','d','e','f','g']

iabcdata :: [Char]
iabcdata = ['g','f','e','d','c','b','a']

shift :: Char -> Int -> Char
shift l n = if (n >= 0)
            then normalShift l n
            else inverseShift l (abs n)

normalShift :: Char -> Int -> Char
normalShift l n = shifter l n abcdata

inverseShift :: Char -> Int -> Char
inverseShift l n = shifter l n reverse(abcdata) -- This is the line

charIdx :: Char -> [Char] -> Int
charIdx target xs = fromJust $ elemIndex target xs

shifter :: Char -> Int -> [Char] -> Char
shifter l n xs = if (n < length (xs))
            then
                picker ((charIdx l xs) + n) xs
            else
                picker ((charIdx l xs) + (n `mod` length (xs))) xs

picker :: Int -> [Char] -> Char
picker n xs = if n < length xs
              then
                xs!!n
              else
                xs!!(n `mod` length (xs))

我的问题是关于这条线的

inverseShift l n = shifter l n reverse(abcdata)

如果我把它改成

inverseShift l n = shifter l n iabcdata

效果很好

另外，当我 reverse(abcdata) == iabcdata 时，它是 True 但是当我在代码中留下 reverse 时，出现以下错误

    * Couldn't match expected type `[Char] -> Char'
                  with actual type `Char'
    * The function `shifter' is applied to four arguments,
      but its type `Char -> Int -> [Char] -> Char' has only three
      In the expression: shifter l n reverse (abcdata)
      In an equation for `inverseShift':
          inverseShift l n = shifter l n reverse (abcdata)
   |
21 | inverseShift l n = shifter l n reverse(abcdata)
   |                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^

    * Couldn't match expected type `[Char]'
                  with actual type `[a0] -> [a0]'
    * Probable cause: `reverse' is applied to too few arguments

我用 reverse(abcdata) 调用 shifter 做错了什么？

Answer 1

括号在 Haskell 中不是这样工作的。按照你写的方式，reverse 和 abcdata 都是 shifter 的参数，但你希望 abcdata 是 reverse 的参数。做 shifter l n (reverse abcdata) 而不是 shifter l n reverse(abcdata).

Answer 2

What am I doing wrong by calling shifter with reverse(abcdata) ?

留言中的答案是：

    * Couldn't match expected type `[Char] -> Char'
                  with actual type `Char'
    * The function `shifter' is applied to four arguments,
      but its type `Char -> Int -> [Char] -> Char' has only three
      In the expression: shifter l n reverse (abcdata)
      In an equation for `inverseShift':
          inverseShift l n = shifter l n reverse (abcdata)

重复，

In an equation for `inverseShift':

inverseShift l n = shifter l n reverse (abcdata)
~~~~~~~~~~~~~ -- mind the gap!

就是 Haskell 解读你的表情的方式。而这个是你写的：

   |
21 | inverseShift l n = shifter l n reverse(abcdata)
   |                    ^^^^^^^^^^^^^^^^^^^^^^^^^^^^

所以你实际上并没有用 reverse(abcdata) 调用 shifter。

您用 reverse 和 (abcdata) （以及 l 和 n）调用它，也有解释在另一个答案中。