Question

有人可以告诉我这里的错误在哪里吗？

let a = [|2;4;6;9;12|];;

a.(0);;
a.(4);;
a.(5);;

let binary_search array size x = 
  let n = size-1 in
  let p = ref 0 in
  let r = ref n in
  while (!p <= !r) do
    let q = (!p + !r)/2;          
    if array.(q) = x
    then raise ((Found_It (q));)                     
    else if (array.(q) <> x) && (array.(q) > x) 
    then (r := q - 1;)            
    else if array.(q) < x 
    then (p := q + 1;)
  done;
  else -1;;

exception Found_It of int;;

如果您对ocaml中的二进制搜索有任何建议，请通知我？

Answer 1

您的问题是您在首先定义之前使用了例外。将exception Found_It ...行移到let binary_search ...行上方。

另外，正如Drew所说，你的问题与Emacs完全无关。

Answer 2

异常Found_It of int ;; （*这里我们声明一个“异常”。这是一个特殊的工具，可以让我们打破循环。）（此特殊异常采用整数作为参数。*）

让binary_search数组大小为x = 设n = size-1 in（*让不可变变量n存储数组的最终索引）让p = ref 0 in（让minpoint q等于0 ）令r = ref n in（让maxpoint r等于n *）

while !p <= !r do     
  let q = ref ((!p + !r)/2) in              (* calculate the midpoint for roughly equal partition *) 
  print_int !q; print_string " "; 
  if array.(!q) = x                (* if x is found at index p *) 
  then raise (Found_It (!q))       (* then break the loop with Found_It, which "carries" the value of q  with it *)           
  else if array.(!q) > x         (* otherwise if q <> x and q > x *)
  then r := !q - 1                 (*change r index to search lower subarray*)
  else p := !q + 1                 (* otherwise if q < x change p index to search upper lower subarra *)
done;                             (* that's the end of the loop                                                *)
-1; 

                       (* so if we reach the end of the loop, output -1 for "Not found"             *)

with Found_It（x） - ＆gt; X;; （*如果我们用q打破了循环，输出q *）

OCaml中的二进制搜索？

2 个答案: