Question

我有一个名为响应的变体类型，它基于另一种变体类型。但我无法使用响应变体类型作为返回值。这是代码：

type ack_error =
  | Not_list        
  | Arg             
  | Password        
  | Permission      
  | Unknown         
  | No_exist        
  | Playlist_max    
  | System          
  | Playlist_load   
  | Update_already  
  | Player_sync     
  | Exist           

type response = Ok | Error of ack_error * int * string * string

(* some code for the function_parse_error_response which signature is : 
 *  str -> (ack_error * int * string * string)
 *)
let parse_response mpd_response =
  if mpd_response = "OK\n" then Ok
  else parse_error_response mpd_response

我创建了一个test.ml：

(* ocamlfind ocamlc -o test -package oUnit -linkpkg -g mpd.ml test.ml *)
open OUnit2

let test_ok test_ctxt =  assert_equal Ok Mpd.parse_response "OK\n"

let mpd_tests =
  "mpd_tests" >:::
    ["test OK" >:: test_ok]

let () =
  run_test_tt_main mpd_tests

当我尝试编译它时，我有以下错误：

ocamlfind ocamlc -o test -package oUnit -linkpkg -g mpd.ml test.ml
File "mpd.ml", line 84, characters 7-40:
Error: This expression has type ack_error * int * string * string
       but an expression was expected of type response

Answer 1

parse_error_response mpd_response的类型为ack_error * int * string * string，而非response。您只需要使用Error构造函数包装它：

let parse_response mpd_response =
  if mpd_response = "OK\n" then Ok
  else Error (parse_error_response mpd_response)

Ocaml：如何使用变量类型作为返回值

1 个答案: