我开始在 ocaml 工作，我无法弄清楚关于 nat 类型的小事

Question

我正在尝试编写类型为 nat -> nat -> nat -> nat 的 function，我将其命名为 sudan_nat 它按照苏丹 function 的原则工作 **

Wikipedia:
   https://en.wikipedia.org/wiki/Sudan_function

//This is the type definition
type nat = Zero | Succ of nat

let add_nat m n =
  let rec helper m = match m with Zero -> n | Succ m -> Succ (helper m)
  in helper m

let zero_nat = Zero
let one_nat  = Succ zero_nat             
let two_nat  = Succ one_nat             
let four_nat  = add_nat two_nat two_nat`



let sudan_nat m n p = 
  let rec helper m n = match m with Zero -> n | Succ m -> Succ (helper m n)
  in helper (add_nat (add_nat m n) p) one_nat

这是我为 sudan_nat 写的代码

我测试 output 的情况是 ((sudan_nat one_nat two_nat two_nat) = (add_nat (add_nat four_nat four_nat) four_nat))

Succ (Succ (Succ (Succ (Succ (Succ Zero)))))

这是左侧的结果

Succ
 (Succ
   (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ Zero)))))))))))

这是右侧的结果我已经摸索了几个小时现在欢迎任何建议谢谢

Answer 1

four_nat是Succ (Succ (Succ (Succ Zero))) 。 那是 4 个Succ构造函数。 你把其中的 3 个加在一起，所以你有... 12。4 乘以 3 = 12。

当谈到评估sudan_nat nat_one nat_two nat_two时，它分解了：

sudan_nat nat_one nat_two nat_two
helper (add_nat (add_nat nat_one nat_two) nat_two) one_nat
helper (add_nat (add_nat Zero nat_two) nat_two) one_nat
helper (add_nat nat_two nat_two) one_nat
helper nat_four one_nat
...
Succ (Succ (Succ (Succ (Succ Zero))))

作为额外的风格建议，您可能希望定义一个模块来组织您的代码。

module Nat =
struct
  type t = Zero | Succ of t
  
  let add m n =
    let rec helper m = 
      match m with 
      | Zero -> n 
      | Succ m -> Succ (helper m)
    in 
    helper m

  let one = Succ Zero
  let two = Succ (Succ Zero)
  let four = add two two
  
  let sudan m n p = 
    let rec helper m n = 
      match m with 
      | Zero -> n 
      | Succ m -> Succ (helper m n)
    in 
    helper (add (add m n) p) one
end