Question

我试图创建自己的字符串shuffle函数：

import System.Random

-- usage case: my_shuffle "something" ""

my_shuffle :: [Char] -> [Char] -> [Char]
my_shuffle [] result = result
my_shuffle s result = do 
    pos <- randomRIO (1, length s)
    my_shuffle (remove_char pos) (result ++ (get_char pos))

get_char :: [Char] -> Int -> Char
get_char s pos  = s !! (pos - 1)

remove_char :: [Char] -> Int -> [Char]
remove_char s pos = take (pos - 1) s ++ drop pos s

它返回错误消息：

substitution_cipher.hs:8:16:
    Couldn't match expected type `[t0]' with actual type `IO a0'
    In the return type of a call of `randomRIO'
    In a stmt of a 'do' expression: pos <- randomRIO (1, length s)
    In the expression:
      do { pos <- randomRIO (1, length s);
           my_shuffle (remove_char pos) (result ++ (get_char pos)) }

我认为它与IO有关，但我不知道如何解决它。

Answer 1

首先，您没有将字符串参数传递给remove_char和get_char。此外，您需要将get_char的结果转换为列表才能使用++。对my_shuffle的递归调用应如下所示：

my_shuffle (remove_char s pos) (result ++ [get_char s pos])

其次，您需要将{mon}用于randomIO，因此my_shuffle的签名应为：

my_shuffle :: [Char] -> [Char] -> IO [Char]

然后最后你需要在基本情况下使用return（因为你需要返回IO [Char]）：

my_shuffle [] result = return result

修复已应用：

import System.Random

my_shuffle :: [Char] -> [Char] -> IO [Char]
my_shuffle [] result = return result
my_shuffle s result = do
     pos <- randomRIO (1, length s)
     my_shuffle (remove_char s pos) (result ++ [get_char s pos])

get_char :: [Char] -> Int -> Char
get_char s pos  = s !! (pos - 1)

remove_char :: [Char] -> Int -> [Char]
remove_char s pos = take (pos - 1) s ++ drop pos s

在字符串shuffle函数中键入错误

1 个答案: