在OCaml中反转int

Question

我正在自学OCaml,我正在练习的主要资源是康奈尔从3110课程中提供的一些问题.其中一个问题是编写一个函数来反转int(即:1234 - > 4321,-1234 - > -4321,2 - > 2,-10 - > -1等).

我有一个有效的解决方案,但我担心它不完全是惯用的OCaml:

let rev_int (i : int) : int =
  let rec power cnt value =
    if value / 10 = 0 then cnt 
    else power (10 * cnt) (value/10) in
  let rec aux pow temp value =
    if value <> 0 then aux (pow/10) (temp + (value mod 10 * pow)) (value / 10)
    else temp in
  aux (power 1 i) 0 i

据我所知,它在所有情况下都能正常工作,但它对我来说似乎是非常"非OCaml",特别是因为我使用两个内部函数运行了两次int的长度.所以我只是想知道是否有更多的"OCaml"方式来做到这一点.

Answer 1

I would say, that the following is idiomatic enough.

(* [rev x] returns such value [y] that its decimal representation
   is a reverse of decimal representation of [x], e.g., 
   [rev 12345 = 54321] *)
let rev n = 
  let rec loop acc n =
    if n = 0 then acc 
    else loop (acc * 10 + n mod 10) (n / 10) in
  loop 0 n

But as Jeffrey said in a comment, your solution is quite idiomatic, although not the nicest one.

Btw, my own style, would be to write like this:

let rev n = 
  let rec loop acc = function
    | 0 -> acc
    | n -> loop (acc * 10 + n mod 10) (n / 10) in 
  loop 0 n

因为我更喜欢模式匹配if/then/else。但这是我个人品味的问题。