OCaml:需要类型为unit Lwt.t的Lwt表达式

Question

我想为我的OUnit测试编写一个小服务器，当客户端通过套接字连接时，向客户端发送数据并退出。

因为我想在测试中使用它，所以我认为应该在线程中创建一个服务器，在线程中创建一个客户端。下面是代码：

open Sys
open Unix
open Lwt

(* a simple server that send a message when a client  connects
 * compile with
 * ocamlfind ocamlc -o lwt_server -package lwt,lwt.unix,unix  -linkpkg -g lwt_server.ml
 * http://baturin.org/code/lwt-counter-server/
 * https://mirage.io/wiki/tutorial-lwt
 * http://aubedesheros.blogspot.fr/2011/05/ocaml-lwt-mini-tutorial.html
 *)

let host = Unix.inet_addr_loopback (* 127.0.0.1 *)
let port = 6600
let max_pending_request = 10

(* Initialize logging capabilities at INFO level *)

let () = Lwt_log.add_rule "*" Lwt_log.Info

let send_message oc =
  Lwt_io.write_line oc "test"

let accept_connection conn =
  let fd, _ = conn in
  let oc = Lwt_io.of_fd Lwt_io.Output fd in
  Lwt.on_failure (send_message oc ) (fun e -> Lwt_log.ign_error (Printexc.to_string e));
  Lwt_log.info "New connection" >>= return

let create_server sock =
  let serve () =
    Lwt_unix.accept sock >>= accept_connection
  in serve

let sock_recv sock maxlen =
  let str = Bytes.create maxlen in
  let recvlen  = recv sock str 0 maxlen [] in
  String.sub str 0 recvlen

let sock_read sock =
  let answer = sock_recv sock 512 in
    Lwt_io.write_line Lwt_io.stdout answer

let create_socket () =
  let sock = Lwt_unix.socket PF_INET SOCK_STREAM 0 in
  Lwt_unix.bind sock @@ ADDR_INET(host, port);
  Lwt_unix.listen sock max_pending_request;
  sock


let () =
  let sock = create_socket () in
    let threads = Lwt.join [create_server sock; sock_read sock] in
      Lwt_main.run threads

下面是我用来编译它的命令和错误消息

ocamlfind ocamlc -o lwt_server -package lwt,lwt.unix,unix  -linkpkg -g lwt_server.ml
File "lwt_server.ml", line 53, characters 28-46:
Error: This expression has type unit -> unit Lwt.t
       but an expression was expected of type unit Lwt.t

第53行是指(不是开玩笑，我已经检查了两次)：

        let threads = Lwt.join [create_server sock; sock_read sock] in

这是我第一次使用Lwt，我不确定它是否是我应该在OUnit中使用的解决方案。

Answer 1

问题是create_server sock返回另一个unit -> unit Lwt.t类型的函数，而不是类型错误中指示的LWT线程(unit Lwt.t)：

let create_server sock =
  let serve () =
    Lwt_unix.accept sock >>= accept_connection
  in serve (* serve is a function, not a thread *)

您可以重写this以返回线程而不是函数来修复this类型错误：

let create_server sock =
  Lwt_unix.accept sock >>= accept_connection

我还没有对代码进行整体测试，但这应该可以修复类型错误。