函数中二分法的运用问题

问题描述：无论k为何值都得出Can't find it.

#include
int binary_search(int arr[],int k,int sz)
{
    int left = 0;
    int right = sz - 1;
    while (left <= right)
    {
        int mid = (left + right) / 2;
        if (arr[mid] < k)
        {
            left = mid + 1;
        }
        else if (arr[mid] > k)
        {
            right = mid - 1;
        }
        else
        {
            return mid;
        }
        return -1;
    }
}
int main()
{
    int arr[] = { 1,2,3,4,5,6,7,8,9,10 };
    int k = 7;
    int sz = sizeof(arr) / sizeof(arr[0]);
    int ret = binary_search(arr, k, sz);
    if (ret == -1)
    {
        printf("Can't find it\n");
    }
    else
    {
        printf("yes,it is:%d\n", ret);
    }
    return 0;
}

代码第22行看一下 return -1的位置。

#include<stdio.h>
int binary_search(int arr[],int k,int sz)
{
    int left = 0;
    int right = sz - 1;
    while (left <= right)
    {
        int mid = (left + right) / 2;
        if (arr[mid] < k)
        {
            left = mid + 1;
        }
        else if (arr[mid] > k)
        {
            right = mid - 1;
        }
        else
        {
            return mid;
        }
    }
    return -1;
}
int main()
{
    int arr[] = { 1,2,3,4,5,6,7,8,9,10 };
    int k = 7;
    int sz = sizeof(arr) / sizeof(arr[0]);
    int ret = binary_search(arr, k, sz);
    if (ret == -1)
    {
        printf("Can't find it\n");
    }
    else
    {
        printf("yes,it is:%d\n", ret);
    }
    return 0;
}