模式匹配 ocaml 中的嵌套元组

Question

>我正在尝试递归浏览文件和文件夹的嵌套列表。当我这样做时，我想打印我访问的文件和文件夹的名称，但我做错了什么。这是我所拥有的：

type 'a fileTree =¬
  | File of 'a¬
  | Folder of 'a * ('a fileTree list)
let printFiles tree =
  let rec visit tree = function
    | [] -> print_string "Done"
    | File f :: t -> visit t
    | Folder (name, contents) :: t -> visit contents
  in
  visit tree

这样做我得到

Error: This expression has type 'a fileTree list -> unit but an expression was expected of type unit

对于第 (File f :: t -> visit t) 行。

我做错了什么？

编辑 1

代码现在是

let printFiles tree =
  let rec visit = function
    | File file -> Printf.printf "file: %sn" file
    | Folder (name, contents) ->
      Printf.printf "folder: %sn" name;
      List.iter visit contents
  in
  List.iter visit tree

但是我仍然在最后一行收到此错误：

Error: This function has type ('a -> unit) -> 'a list -> unit
   It is applied to too many arguments; maybe you forgot a `;'

Answer 1

你忘记了这里的括号：

print_string "File " ^ f

它被评估为 (print_string "File ") ^ f ，但您期望的是

print_string ("File " ^ f)

"Folder"情况也是如此。

更新

在更新的示例中，您有一个将两个参数应用于一个参数的函数。我想，而不是：

 let rec visit tree = function

你想写

 let rec visit = function

相当于

 let rec visit tree = match tree with

Answer 2

问题是你已经将visit写为列表上的函数，但在文件树上调用了它。

让我们将其更改为树上的函数，并使用List.iter来遍历列表：

let print_files tree =
  let rec visit = function
    | File file -> 
      Printf.printf "file: %sn" file
    | Folder (name, contents) ->
      Printf.printf "folder: %sn" name;
      List.iter visit contents in
  visit tree

Answer 3

最后我通过以下方式解决了它：

let writeFilesFromTree tree =
  let rec visit = function
    | [] -> print_string "-n"
    | File f :: t -> Printf.printf "file: %sn" f ; visit t
    | Folder (name, contents) :: t ->
      Printf.printf "name: %sn" name ;
      visit contents ;
      visit t in
  visit tree;;

它仍然感觉根本不是一个最佳解决方案，而且我感觉我正在通过最后两次连续调用visit来破坏尾递归。