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此表达式的类型为unit,但表达式的类型应为'a Client.io

ocaml

0

Vlam · 技术社区 · 8 月前

我试图在下面用OCaml编写一个简单的可执行代码。

open Printf
open Lwt
open Cohttp
open Cohttp_lwt_unix
open Yojson

let () =
  let ip = "8.8.8.8" in
  let key = "" in
  let uri =
    Uri.of_string
      ("https://api.ip2location.io/?format=json&key=" ^ key ^ "&ip=" ^ ip)
  in
  Lwt_main.run
    ( Client.get uri >>= fun (resp, body) ->
      let code = resp |> Response.status |> Code.code_of_status in
      let json_promise = body |> Cohttp_lwt.Body.to_string in
      json_promise >>= fun json_string ->
      let json = Basic.from_string json_string in
      let open Yojson.Basic.Util in
      if code == 200 then
        if member "usage_type" json <> `Null then
          let usage_type = json |> member "usage_type" |> to_string in
          printf "usage_type: %s\n" usage_type
        else
          printf
            "ERROR: The usage_type field requires a paid subscription to the \
             Starter plan or higher."
      else if (code == 400 || code == 401) && member "error" json <> `Null then
        let error_message =
          json |> member "error" |> member "error_message" |> to_string
        in
        printf "ERROR: " ^ error_message
      else printf "HTTP Code: " ^ Int.to_string code )

但我跑步时总是看到下面 dune build .

File "bin/main.ml", line 24, characters 10-46:
24 |           printf "usage_type: %s\n" usage_type
               ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Error: This expression has type unit but an expression was expected of type
         'a Client.io

从其他StackOverflow帖子来看,这似乎与 if/else 但我已经确保了所有 if/else 正在使用 printf .

如果有人能告诉我我还做错了什么,我将不胜感激。

2 回复 | 直到 8 月前

1

Chris 8 月前

Andreas Rossberg发现了一个问题,但你也看到了一些 operator precedence 问题。

这类事情:

printf "ERROR: " ^ error_message

还有这个:

printf "HTTP Code: " ^ Int.to_string code

看起来很可疑,因为 printf "HTTP code: " 评估为 () 无法提供给 ^ 。看起来你是想写这样的东西:

printf ("HTTP Code: " ^ Int.to_string code)

但你也可以利用 printf 编写以下内容,它更清晰,没有运算符优先级问题。

printf "HTTP Code: %d" code

还要小心打开几个模块。如果多个运算符定义了相同的运算符,或者它们重新定义了标准库运算符,则可能会出现一些意外行为。

除了完全符合条件的名字,你还可以当地开放模块,缓解了这个问题。但是,由于您已经在代码中使用过一次,因此您似乎知道这个选项。

从风格上讲,我可能只是重构了这个:

        let error_message =
          json |> member "error" |> member "error_message" |> to_string
        in
        printf "ERROR: " ^ error_message

致:

        json 
        |> member "error" 
        |> member "error_message" 
        |> to_string
        |> printf "ERROR: %s"

也有意见,但有机会返工:

      if code == 200 then
        if member "usage_type" json <> `Null then
          let usage_type = json |> member "usage_type" |> to_string in
          printf "usage_type: %s\n" usage_type
        else
          printf
            "ERROR: The usage_type field requires a paid subscription to the \
             Starter plan or higher."
      else if (code == 400 || code == 401) && member "error" json <> > `Null then
        let error_message =
          json |> member "error" |> member "error_message" |> to_string
        in
        printf "ERROR: " ^ error_message
      else printf "HTTP Code: " ^ Int.to_string code )

使用模式匹配和条件保护:

      match code with
      | 200 when member "usage_type" json <> `Null ->
        json
        |> member "usage_type" 
        |> to_string
        |> printf "usage_type: %s\n"
      | 200 ->
        printf
          "ERROR: The usage_type field requires a paid subscription to the \
           Starter plan or higher."
      | (400 | 401) when member "error" json <> `Null ->
        json 
        |> member "error" 
        |> member "error_message" 
        |> to_string
        |> printf "ERROR: %s"
      | _ -> printf "HTTP Code: %d" code

2

1

Andreas Rossberg 8 月前

我怀疑这是由于 >>= ,其Lwt类型为

(>>=) : 'a Lwt.t -> ('a -> 'b Lwt.t) -> 'b Lwt.t

在左侧,您将传递以下结果 Client.get url ,因此右侧必须是返回相同类型的函数。同样地, Lwt.run 期待A Lwt.t (我对所有涉及的图书馆都不太熟悉,但 Client.io 大概是一个类型同义词扩展到 Lwt。t 这里)

这个例子的修复应该是使用 >|= 而不是第二个 >>= ,其类型为

(>|=) : 'a Lwt.t -> ('a -> 'b) -> 'b Lwt.t