为什么口译员告诉我 "This kind of expression is not allowed as right-hand side of `let rec'"

Question

我编写了一个 ocaml 程序，通过解析器组合器解析算术表达式。

type 'a parser = char list -> ('a * (char list)) list
let return (x: 'a): 'a parser = fun input -> [x, input]
let fail: 'a parser = fun _ -> []
let ( >>= ) (p: 'a parser) (f : 'a -> 'b parser): 'b parser =
    fun input -> List.map (fun (x, i) -> f x i) (p input) |> List.flatten

let ( ||| ) (p: 'a parser) (q: 'a parser) = 
    fun input -> (p input) @ (q input)

let token: (char parser) = function
    | x::xs -> [x, xs]
    | [] -> []

let char(c: char): (char parser) =
    token >>= fun x ->
    if x = c then return x else fail

let digit: (char parser) =
    token >>= fun x ->
    if x >= '0' && x <= '9' then return x else fail

let rec many(p: 'a parser): 'a list parser =
    (p >>= fun x ->
    many p >>= fun xs ->
    (return (x::xs))) 
    ||| (return [])

let number = 
    many digit >>= fun x -> 
    return (List.fold_left (fun l r -> l * 10 + (int_of_char r - int_of_char '0')) 0 x)

type expr = Add of (expr * expr)
            | Sub of (expr * expr)
            | Mul of (expr * expr)
            | Div of (expr * expr)
            | Neg of expr

上面的代码运行良好。

let rec expression: expr parser =
    term >>= fun l ->
    (char '+' ||| char '-') >>= fun op ->
    term >>= fun r ->
    if op = '+' then return (Add (l, r)) else return (Sub (l, r))
and term: expr parser =
    factor >>= fun l ->
    (char '*' ||| char '/') >>= fun op ->
    factor >>= fun r ->
    if op = '*' then return (Mul (l, r)) else return (Div (l, r))
and factor: expr parser =
    expression
    ||| (char '(' >>= fun _ ->
        expression >>= fun e ->
        char ')' >>= fun _ ->
        return e)
    ||| (char '-' >>= fun _ ->
        factor >>= fun e -> 
        return (Neg e))

但是这段代码有错误

    Error: This kind of expression is not allowed as right-hand side of `let rec'

本页（link）说： "the compiler only allows three possible constructs to show up on the righthand side of a let rec: a function definition, a constructor, or the lazy keyword."

我认为类型解析器是一个函数而不是一个值，为什么会这样？

Answer 1

"Function definition" 表示文字 fun x -> 结构。函数式语言考虑部分函数应用程序，例如

factor >>= fun l -> ...

redexes，而不是文字值，这意味着当它们出现在简单变量绑定的 RHS 上时，严格的语言必须立即对其求值，例如

term = factor >>= fun l -> ...

由于通常急于计算递归定义的 RHS 会产生无限循环，因此严格的语言通常要么禁止该构造，要么要求将绑定显式标记为惰性。

Answer 2

右边应该"function definition"不只是一个函数。函数定义是手册form fun -> or function ..., including syntactic sugar for let f x .... A more correct wording is in sections 6.7.1 and 7.3的句法结构。

为什么口译员告诉我 "This kind of expression is not allowed as right-hand side of `let rec'"

why the interpreter tell me "This kind of expression is not allowed as right-hand side of `let rec'"

ocaml

functional-programming