SML如何在char列表上返回int？

发布于 2024-09-19 03:51:56 字数 776 浏览 1 评论 0原文

我有点难以弄清楚如何将每个处理后的字符恢复为 int 值。

该函数应该像这样工作： val caesar = fn : int * int -> int

所以如果 k = 2466 且 n = 2，那么输出应该是 4688

希望代码不会太奇怪（我是 SML 新手）。

 (* Load Libs *)
    load "Int";
    load "Real";
    load "String";
    load "Char";
    load "List";


    fun caesar (k, n) =
      let
        fun k_string (i) = Int.toString(i)
        fun item_k_char (x, y) = Char.ord (List.nth (x, y))

        val val_k_string = k_string(k)
        val k_explode = String.explode(val_k_string)
        val counter = ref 0
        val counter_end = (String.size(val_k_string) - 1)

      in 
        while (!counter >= counter_end) do (
          item_k_char(k_explode, !counter) + n;
          counter := !counter + 1
        )
      end;

原文

I'm having a bit difficulty figuring out, how to get each of the processed chars back to an int value.

The function should work like: val caesar = fn : int * int -> int

So if k = 2466 and n = 2, then the output should be 4688

Hope the code isn't too weird (I'm a SML newbie).

 (* Load Libs *)
    load "Int";
    load "Real";
    load "String";
    load "Char";
    load "List";


    fun caesar (k, n) =
      let
        fun k_string (i) = Int.toString(i)
        fun item_k_char (x, y) = Char.ord (List.nth (x, y))

        val val_k_string = k_string(k)
        val k_explode = String.explode(val_k_string)
        val counter = ref 0
        val counter_end = (String.size(val_k_string) - 1)

      in 
        while (!counter >= counter_end) do (
          item_k_char(k_explode, !counter) + n;
          counter := !counter + 1
        )
      end;

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习惯那些不曾习惯的习惯 2024-09-26 03:51:57

while 循环并不是最好的工具。相反，您可以使用 map 它为给定列表中的每个项目执行给定函数，并返回一个包含每次调用该函数的结果的新列表。

换句话说： map (fn c => (Char.ord c) + 2) [#"2", #"4", #"6", #"6"] 将返回[52,54,56,56]。您可以将其转换回 char 并使用 String.implode 来获取字符串“4688”。

您可能还想添加一些逻辑，使数字“环绕”，即 caesar (7,7) 变为 4。

所以总而言之你的代码变成：

fun caesar (k, n) =
  let
    val val_k_string = Int.toString k
    val k_explode = String.explode val_k_string
    val ints = map (fn c => (Char.ord c) + n) k_explode
    val wrappedAroundInts = map (fn i => (i - 48) mod 10 + 48) ints
    val chars = map Char.chr wrappedAroundInts
    val string = String.implode chars
  in
    Option.valOf (Int.fromString string)
  end

A while loop isn't the best tool here. Instead you can use map which executes a given function for each item in a given list and returns a new list containing the result of each call to the function.

In other words: map (fn c => (Char.ord c) + 2) [#"2", #"4", #"6", #"6"] will return [52,54,56,56]. You can the convert this back to char and use String.implode to get the string "4688".

You probably also want to add some logic so that the numbers "wrap around", i.e. caesar (7,7) becomes 4.

So all in all your code becomes:

fun caesar (k, n) =
  let
    val val_k_string = Int.toString k
    val k_explode = String.explode val_k_string
    val ints = map (fn c => (Char.ord c) + n) k_explode
    val wrappedAroundInts = map (fn i => (i - 48) mod 10 + 48) ints
    val chars = map Char.chr wrappedAroundInts
    val string = String.implode chars
  in
    Option.valOf (Int.fromString string)
  end

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