如何简化这个 ocaml 模式匹配代码?

发布于 2024-08-13 04:49:48 字数 1864 浏览 6 评论 0原文

我正在编写一个简单的小 ocaml 程序,它从文件中读取代数语句,使用 ocamllex/ocamlyacc 将其解析为 AST,减少它,然后打印它。我减少表情的部分似乎有点……丑陋。我有什么办法可以简化它吗?

(* ocaml doesn't seem to be able to take arithmetic operators
 as functions, so define these wrappers for them *)
let add x y =
  x + y

let sub x y =
  x - y

let mul x y =
  x * y

let div x y =
  x / y

(* Are term1 and term2 both ints? *)
let both_ints term1 term2 =
  match (term1, term2) with
    | (Term (Number x), Term (Number y)) -> true
    | (_, _) -> false

(* We know that both terms are reducable to numbers, so combine
  them *)
let combine_terms func x y =
  match (x, y) with
      (Term (Number t1), Term (Number t2)) ->
        (Term (Number (func  t1  t2)))
    | (_, _) -> raise InvalidArg

(* Reduce the expression as much as possible *)
let rec reduce_expr expr =
  match expr with
      Plus (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms add reduced_x reduced_y)
      else
        Plus (reduced_x, reduced_y)
    | Minus (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms sub reduced_x reduced_y)
      else
        Minus (reduced_x, reduced_y)
    | Multiply (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms mul reduced_x reduced_y)
      else
        Multiply (reduced_x, reduced_y)
    | Divide (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms div reduced_x reduced_y)
      else
        Divide (reduced_x, reduced_y)
    | Term x -> Term x

I'm writing a simple little ocaml program that reads an algebraic statement in from a file, parses it into an AST using ocamllex/ocamlyacc, reduces it, and then prints it. The part where I'm reducing the expression seems a bit... ugly. Is there any way I can simplify it?

(* ocaml doesn't seem to be able to take arithmetic operators
 as functions, so define these wrappers for them *)
let add x y =
  x + y

let sub x y =
  x - y

let mul x y =
  x * y

let div x y =
  x / y

(* Are term1 and term2 both ints? *)
let both_ints term1 term2 =
  match (term1, term2) with
    | (Term (Number x), Term (Number y)) -> true
    | (_, _) -> false

(* We know that both terms are reducable to numbers, so combine
  them *)
let combine_terms func x y =
  match (x, y) with
      (Term (Number t1), Term (Number t2)) ->
        (Term (Number (func  t1  t2)))
    | (_, _) -> raise InvalidArg

(* Reduce the expression as much as possible *)
let rec reduce_expr expr =
  match expr with
      Plus (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms add reduced_x reduced_y)
      else
        Plus (reduced_x, reduced_y)
    | Minus (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms sub reduced_x reduced_y)
      else
        Minus (reduced_x, reduced_y)
    | Multiply (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms mul reduced_x reduced_y)
      else
        Multiply (reduced_x, reduced_y)
    | Divide (x, y) ->
        let reduced_x = reduce_expr x
        and reduced_y = reduce_expr y in
      if both_ints reduced_x reduced_y then
        (combine_terms div reduced_x reduced_y)
      else
        Divide (reduced_x, reduced_y)
    | Term x -> Term x

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城歌 2024-08-20 04:49:48

您可以通过匹配包含 Number 的类型来删除 both_intscombine_terms 函数以及一些 if 语句在操作功能内,例如:

let div x y = match x,y with
    | Number x, Number y -> Number (x / y)
    | _ -> Divide (x,y)

...
let rec reduce_expr expr = match expr with
    ...
    | Divide (x,y) -> div (reduce_expr x) (reduce_expr y)
    ...

You can remove the both_ints, and combine_terms functions as well as some if statements, by matching the type containing Number within the operation functions, such as:

let div x y = match x,y with
    | Number x, Number y -> Number (x / y)
    | _ -> Divide (x,y)

...
let rec reduce_expr expr = match expr with
    ...
    | Divide (x,y) -> div (reduce_expr x) (reduce_expr y)
    ...
~没有更多了~
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