Types OCaml期望的模式不';在模式匹配过程中,任何位置都不会出现

Types OCaml期望的模式不';在模式匹配过程中,任何位置都不会出现,types,pattern-matching,ocaml,Types,Pattern Matching,Ocaml,线路 Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] -> 提出错误 37 | Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] -> ^^^^^^^^^^^^^ Error: This pattern

线路

Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] ->
提出错误

37 |   Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] ->
       ^^^^^^^^^^^^^
Error: This pattern matches values of type regexp_t
       but a pattern was expected which matches values of type
         int list * int list
33 |   Empty_String -> let x = fresh() and y = fresh() in
       ^^^^^^^^^^^^
Error: This variant pattern is expected to have type int list
       The constructor Empty_String does not belong to type list
当a、b显然需要regexp\u to\u nfa函数使用类型regexp\u t时,为什么编译器会认为int list*int list需要匹配

带有类似错误的问题表明错误源于前面的表达式,其类型为int list*int list。但是,如果我将Concat行移到开头(在空字符串->之前),我会得到错误

37 |   Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] ->
       ^^^^^^^^^^^^^
Error: This pattern matches values of type regexp_t
       but a pattern was expected which matches values of type
         int list * int list
33 |   Empty_String -> let x = fresh() and y = fresh() in
       ^^^^^^^^^^^^
Error: This variant pattern is expected to have type int list
       The constructor Empty_String does not belong to type list
,这表明Concat((a,b))或使用它的行中的某物具有int-list类型,这就没有什么意义了

作为参考,以下是我正在使用的类型和我正在尝试调试的函数的代码:

type ('q, 's) transition = 'q * 's option * 'q
type ('q, 's) nfa_t = {
    sigma : 's list;
    qs : 'q list;
    q0 : 'q;
    fs : 'q list;
    delta : ('q, 's) transition list;
}

type regexp_t =
  | Empty_String
  | Char of char
  | Union of regexp_t * regexp_t
  | Concat of regexp_t * regexp_t
  | Star of regexp_t

let fresh =
  let cntr = ref 0 in
  fun () ->
    cntr := !cntr + 1 ;
    !cntr

let rec regexp_to_nfa (regexp: regexp_t) : (int, char) nfa_t = match regexp with
  Empty_String -> let x = fresh() and y = fresh() in 
  {sigma = []; qs = [x;y]; q0 = x; fs = [y]; delta = []}|
  Char(a) -> let x = fresh() and y = fresh() in 
  {sigma = [a]; qs = [x;y]; q0 = x; fs = [y]; delta = [(x,Some a,y)]}|
  Union((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b and r = fresh() and s = fresh() in match (x.fs, y.fs) with ([t],[u]) -> 
  {sigma = Sets.union x.sigma y.sigma; qs = Sets.union (Sets.union x.qs y.qs) [r;s]; q0 = r; fs = [s]; 
  delta = Sets.union (Sets.union x.delta y.delta) [(r, None, x.q0); (r,None,y.q0); (t,None,s); (u,None,s)]}|
  Concat((a,b)) -> let x = regexp_to_nfa a and y = regexp_to_nfa b in match x.fs with [t] -> 
  {sigma = Sets.union x.sigma y.sigma; qs = Sets.union x.qs y.qs; q0 = x.q0; fs = y.fs; 
  delta = Sets.union (Sets.union x.delta y.delta) [(t,None,y.q0)]} |
  Star(a) -> let x = regexp_to_nfa a and r = fresh() and s = fresh() in match x.fs with [t] -> 
  {sigma = x.sigma; qs = Sets.union x.qs [r;s]; q0 = r; fs = [s]; 
  delta = Sets.union x.delta [(r, None, x.q0);(r,None,s);(s,None,r);(t, None, r)]}

您有一个嵌套的
匹配项
。因此,
Concat
模式被解析为嵌套匹配的一部分

您需要使用
开始/结束
或括号来分隔嵌套在另一个
匹配
中的
匹配

如果继续使用嵌套匹配,还需要处理
(x.fs,y.fs)
没有预期形式的情况(即,其中一个列表的长度不是1)。如果你确定它们从来都不是空的,你可以这样重写:

Union((a,b)) ->
    let x = regexp_to_nfa a
    and y = regexp_to_nfa b
    and r = fresh()
    and s = fresh() in
    { sigma = Sets.union x.sigma y.sigma;
      qs = Sets.union (Sets.union x.qs y.qs) [r;s];
      q0 = r;
      fs = [s]; 
      delta = Sets.union (Sets.union x.delta y.delta)
                    [(r, None, x.q0);
                     (r, None, y.q0);
                     (List.hd x.fs, None, s);
                     (List.hd y.fs, None, s)]
    }|
Concat((a,b)) -> . . .

(值得一提的是,您使用的是一种非常密集的编码风格。最好是更松散地格式化,甚至使用格式化工具来自动排列代码。)

关于使用List.hd的好主意。递归应确保x.fs、y.fs的大小始终正好为1,我的原始匹配语句试图利用这一点。但是OCaml穷举性检查(通常非常棒)不知道您知道什么。因此,您可能会收到警告,或者必须插入注释(这些注释很难看),才能关闭警告。