OCaml中列表上的相互递归语法错误

我使用的是OCAMLV4.00.1。我正在尝试使用相互递归编写一个函数，以获取一个列表并返回一个int。int是获取列表中的交替元素并将它们相互相加和相减的结果。例如，列表[1；2；3；4]将导致1+2-3+4=4

我的代码如下：

let alt list =
  let rec add xs = match xs with 
    [] -> 0 
    | x::xs -> x + (sub xs)
  and sub xs = match xs with 
    [] -> 0
    | x::xs -> x - (add xs);;

---

OCaml在；；上抛出语法错误；；在代码的最后。我不确定从何处开始找出这个错误的真正原因。

我怀疑您忘了将

添加到
let
binding–粗体中的
部分

let alt list =
  let rec add xs =
    match xs with 
      | [] -> 0 
      | x::xs -> x + (sub xs)
  and sub xs =
    match xs with 
      | [] -> 0
      | x::xs -> x - (add xs)
  in
  add list
我们可以为
函数
交换
匹配
语法，在这里提高了可读性

let alt list =
  let rec add = function
    | [] -> 0
    | x::xs -> x + (sub xs)
  and sub = function
    | [] -> 0
    | x::xs -> x - (add xs)
  in
  add list

我怀疑您忘了在
let
binding中添加
。
部分–更改粗体

let alt list =
  let rec add xs =
    match xs with 
      | [] -> 0 
      | x::xs -> x + (sub xs)
  and sub xs =
    match xs with 
      | [] -> 0
      | x::xs -> x - (add xs)
  in
  add list
我们可以为
函数
交换
匹配
语法，在这里提高了可读性

let alt list =
  let rec add = function
    | [] -> 0
    | x::xs -> x + (sub xs)
  and sub = function
    | [] -> 0
    | x::xs -> x - (add xs)
  in
  add list

我想你是对的！我今晚到家后再查。非常感谢。这很有效，谢谢你！我在哪里可以读到关于使用函数而不是匹配语法的内容？我想你是对的！我今晚到家后再查。非常感谢。这很有效，谢谢你！在哪里可以阅读有关使用函数而不是匹配语法的内容？