Question

我在学校学习OCaml，最近遇到了一个我无法理解的作业计划，并希望有人可以向我解释。这是代码：

(* Natural numbers can be defined as follows:
   type = ZERO | SUCC of nat;;

   For instance, two = SUCC (SUCC ZERO) and three = SUCC (SUCC (SUCC ZERO)).

  Write a function 'natadd' that adds two natural numbers in this fashion. *)

# type = ZERO | SUCC of nat;;
# let two = SUCC (SUCC ZERO);;
# let three = SUCC (SUCC (SUCC ZERO));;
# let rec natadd : nat -> nat -> nat =
    fun n1 n2 ->
      match n1 with
      | ZERO -> n2
      | SUCC n1 -> SUCC (natadd n1 n2)
  ;;

这是代码的示例输出：

# natadd two three;;
- : nat = SUCC (SUCC (SUCC (SUCC (SUCC ZERO))))

我不明白的是匹配声明。这是否意味着如果n1非零，那么它会添加SUCC并使用[SUCC n1]作为新参数代替n1进行递归调用？

Answer 1

不，它不使用SUCC n1作为递归调用的参数，它仅使用n1（匹配的SUCC的一部分）作为参数。然后将SUCC应用于递归调用的结果。

代码可能有点令人困惑，因为有两个名为n1的变量。更好的可能是

let rec natadd : nat -> nat -> nat = fun a b ->
  match a with
  | ZERO -> b
  | SUCC a_pred -> SUCC (natadd a_pred b)

OCaml添加自然数字

1 个答案: