带有字符列表的 Haskell 递归

Question

我正在研究 Haskell 并且出现了一些问题，我被困了几天......所以我基本上试图通过字符列表（字符串）进行递归，一旦它到达字符串的末尾，有没有办法让我再次从头开始递归列表吗？

这是用于递归遍历字符串一次的代码。

repeat2' :: [Char] -> Int -> [Char]
repeat2' [] _ = undefined
repeat2' (x:xs) 1 = [x]
repeat2' (x:xs) k
    | (k > 0) = x: repeat2' xs(k-1)

它适用于

repeat2' "嘿" 1 = "h"

repeat2' "嘿" 2 = "他"

repeat2' “嘿” 3 = “嘿”

但是一旦我尝试了

repeat2' "hey" 4 = "hey*** 例外：Prelude.undefined"

因为它会“repeat2' [] _ = undefined”;

但我希望它打印出来

repeat2' “嘿” 4 = “嘿”

..那么我该如何回到字符串的头部呢？

感谢您的帮助！

Answer 1

让我标记方程式以便稍后引用它们。

   repeat2' :: [Char] -> Int -> [Char]
1) repeat2' [] _ = undefined
2) repeat2' (x:xs) 1 = [x]
   repeat2' (x:xs) k
3)     | (k > 0) = x: repeat2' xs(k-1)

让我们追踪repeat2' "hey" 4 的减少。首先我会将字符串脱糖成一个列表

repeat2' ['h', 'e', 'y'] 4

然后将列表放入其缺点单元格

repeat2' ('h':'e':'y':[]) 4

现在我们可以开始了

repeat2' ('h':'e':'y':[]) 4
-- equation 3 applies, x = 'h', xs = 'e':'y':[], k = 4
'h' : repeat2' ('e':'y':[]) 3
-- the first element of the result is known now, so it is printed by the repl
-- equation 3 applies ot the remainder
'h' : 'e' : repeat2' ('y':[]) 2
-- e is now known and printed
-- equation 3 applies
'h' : 'e' : 'y' : repeat2' [] 1 
-- y is known and printed
-- equation 1 applies
'h' : 'e' : 'y' : undefined
-- the repl errors out because of undefined

从而解释输出

"hey***Exception: Prelude.undefined

希望这能帮助你理解。

一个问题是，为了循环列表，您需要保留它以进行递归调用。请注意，当我们打印 h、e、y 时，接下来要使用的信息不再存在。除了您正在遍历的列表之外，您还需要原封不动地传递列表。