Ошибки Реализовать переводчик в OCaml

Question

я пытаюсь написать интерпретатор в OCaml, но я не знаю, устранить ошибку в этой программе:

SYNTAX 

type ide = string 
type exp = 
    | Eint of int 
    | Ebool of bool 
    | Den of ide 
    | Sum of exp * exp 
    | Diff of exp * exp 
    | Prod of exp * exp 
    | Eq of exp * exp 
    | Minus of exp 
    | Iszero of exp 
    | Or of exp * exp 
    | And of exp * exp 
    | Not of exp 
    | Ifthenelse of exp * exp * exp 
    | Let of ide * exp * exp 
    | Fun of ide * exp 
    | Apply of exp * exp 
    | Letrec of ide * ide * exp * exp 
    | Etup of tuple (*Tupla come espressione*) 
    | Pipe of tuple (*Concatenazione di funzioni*) 
    | ManyTimes of int * exp (*Esecuzione iterata di una funzione*) 
and tuple = 
    | Nil (*Tupla vuota*) 
    | Seq of exp * tuple (*Tupla di espressioni*) 
;; 

SEMANTIC 

type eval= 
    | Int of int 
    | Bool of bool 
    | Unbound 
    | RecFunVal of ide * ide * exp * eval env 
    | Funval of efun 
    | ValTup of etuple 
and efun = ide * exp * eval env 
and etuple = 
    | Nil 
    | Seq of eval * etuple 
;; 

RUN-TIME SUPPORT 

     | Fun(i,a) -> Funval(i,a,r) 
     | Letrec(f, i, fBody, letBody) -> 
      let benv = bind(r, f, (RecFunVal(f, i, fBody, r))) 
      in sem(letBody, benv) 
     | Etup(tup) -> (match tup with 
      | Seq(ex1, tupla) -> 
       let evex1 = sem(ex1, r) in 
       let ValTup(etupla) = sem(Etup(tupla), r) in 
        ValTup(Seq(evex1, etupla)) 
      | Nil -> ValTup(Nil)) 
     | Apply(Den f, arg1) -> 
      (let fclosure= sem(Den f, r) in 
       match fclosure with 
       | Funval(arg, fbody, fDecEnv) -> 
        sem(fbody, bind(fDecEnv, arg, sem(arg1, r))) 
       | RecFunVal(f, arg, fbody, fDecEnv) -> 
        let aVal= sem(arg1, r) in 
        let rEnv= bind(fDecEnv, f, fclosure) in 
        let aEnv= bind(rEnv, arg, aVal) in 
         sem(fbody, aEnv) 
       | _ -> failwith("non functional value")) 
     | Apply(Pipe tup, arg) -> applyPipe tup arg r 
     | Apply(_,_) -> failwith("not function") 

    and applyPipe tup argo r = match tup with 
     | Seq(Den f, tupla) -> 
       applyPipe tupla (Apply(Den f,argo)) r 
     | Seq(Pipe(tuplaP),tupla) -> 
      let appf = applyPipe tuplaP argo r in 
       applyPipe tupla appf r     (**) 
     | Nil -> sem(argo,r) 
     | _ -> failwith("Not a valid Pipe") 
    ;; 

Ошибка на линии (***): «тип типа кортежа не имеет конструктора« Труба » Как его решить?

Answer 1

Компилятор ожидает, что первый аргумент applyPipe будет иметь тип tuple. На линии (***) значение applyPipe применяется к значению Pipe(tupla), которое имеет тип exp.