我有一个 Ocaml 函数的问题，它接受两个输入并且应该给出一个特定的输出

Question

我正在处理家庭作业问题，通常我知道为什么会出现错误，但是，对于这两个问题，我不知道发生了什么。我不确定是模式匹配问题还是模式匹配后我试图解决问题的方式。

Given a STRING LIST and INT return a STRING LIST 
that contains all the strings with a length greater than the given int

    let rec long_strings (l: string list) (i: int) : string list = 
            match l with
            | [] -> []
            | head :: tail -> if (String.length head > i) 
                                 then head :: (long_strings tail) 
                                 else (long_strings tail)

错误

This expression has type int -> string list
but an expression was expected of type string list

另一个类似的问题...

(*Given an INT LIST and INT divide the numbers in the given int 
 list by the given int and return an (INT * INT) LIST containing 
  the dividend and the remainder*)
let rec remainders (l: int list) (i: int) : (int * int) list = 
        match l with
        | [] -> []
        | head :: tail -> ((head / i), (head mod num)) :: (remainders tail)
;;

错误

This expression has type int -> (int * int) list
but an expression was expected of type (int * int) list

Answer 1

您缺少一个参数。 long_strings 的类型是 string list -> int -> string list。因此，如果您将 long_strings 应用于字符串列表，您将获得一个函数：int -> string list。例如，可以定义

let list = ["1"; "12"; "123"]
let filter_list = long_strings list
let _ = assert ( filter_list 0 = ["1"; "12"; "123"] )
let _ = assert ( filter_list 2 = ["123"] )

然而，在

head :: long_strings tail

您期望 long_strings tail 是一个列表。因此错误消息

This expression has type int -> string list but an expression was expected of type string list

告诉你 long_strings tail 不是一个列表，而是一个接受 int 和 returns 和 string_list.

的函数

换句话说，您忘记了 i 参数。

我有一个 Ocaml 函数的问题，它接受两个输入并且应该给出一个特定的输出

I am having an issue with an Ocaml function that takes in two inputs and is supposed ti give a specific output

ocaml