为什么在比较线性搜索和二进制搜索时每次都为零？

Question

我想通过线性和二分搜索找出在数组中查找元素所用的时间。我已经传递了变量 n ，它接受一个整数，并且它不断将它的值从 1 减小到 0。我想获得用于比较线性和二进制搜索的总时间表图表，即从数组大小 1 到任何大数。但是对于这些案例中的每一个，我都得到了 0.000000 次。

我的目标是绘制当 (x) 即输入大小为 0 到任何大数时这两个搜索所经过的时间。

#include<stdio.h>
#include<stdlib.h>
#include<time.h>

long int lin_search(long int [], long int, long int );
long int bin_Search(long int [], long int, long int, long int);
void merge(long int [], long int, long int, long int );
void mergeSort(long int [], long int, long int);

long int main()
{
    long int n, *arr, x;
    clock_t start, end;
    FILE *fp;
    double lin_time , bin_time ;

    //Destination file:  "search_comp.txt"
    fp = fopen("search_comp.txt", "a");

    printf("\nEnter total no of inputs:\t");
    scanf("%ld", &n);
    printf("\n::::::::::::::::::::::::::::\n");
    while (n--)
    {
        arr = (long int *)malloc(n * sizeof(long int));
        for (long int i = 0; i < n; i++)
        {
            arr[i] = rand() % n;
        }

        mergeSort(arr, 0, n - 1);
        x = rand() % n;
        //Linear search time..
        start = clock();
        long int res1 = lin_search(arr, n, x);
        end = clock();
        lin_time = (double)(end - start) / (double)(CLOCKS_PER_SEC);

        //Binary search time ...
        start = clock();
        long int res2 = bin_Search(arr, 0, n - 1, x);
        end = clock();
        bin_time = (double)(end - start) / (double)(CLOCKS_PER_SEC);

        //File creation data taking input........
        fprintf(fp, "%ld;%lf;%lf;%ld;%ld\n", n, lin_time, bin_time, res1, res2);
        free(arr);
    }

    printf("\nTask Completed!\n");
    fclose(fp);
    return 0;
}

long int lin_search(long int arr[],long int n,long int x)
{
    long int i;
    for (i = 0; i < n; i++)
    {
        if (arr[i] == x)
        {
            return i;
        }
    }
    return -1;
}

long int bin_Search(long int arr[],long int l,long int r,long int x)
{
    while (l <= r)
    {
        long int m = l + (r - l) / 2;

        if (arr[m] == x)
            return m;

        if (arr[m] < x)
            l = m + 1;

        else
            r = m - 1;
    }
    return -1;
}

void merge(long int arr[],long int l,long int m,long int r)
{
    long int i, j, k;
    long int n1 = m - l + 1;
    long int n2 = r - m;

    long int L[n1], R[n2];

    for (i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for (j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];


    i = 0; 
    j = 0; 
    k = l; 
    while (i < n1 && j < n2)
    {
        if (L[i] <= R[j])
        {
            arr[k] = L[i];
            i++;
        }
        else
        {
            arr[k] = R[j];
            j++;
        }
        k++;
    }

    while (i < n1)
    {
        arr[k] = L[i];
        i++;
        k++;
    }

    while (j < n2)
    {
        arr[k] = R[j];
        j++;
        k++;
    }
}

void mergeSort(long int arr[],long int l,long int r)
{
    if (l < r)
    {

        long int m = l + (r - l) / 2;

        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);
        merge(arr, l, m, r);
    }
}

此外，如果我省略 while 循环（最多 100,000 个），即程序运行良好

while(n--)
{
    :::::
    :::::
    :::::
}
printf("\nTask Completed!\n");
fclose(fp);
return 0;

对于较高的 n 值，程序不会显示我的 Task Completed! 消息。

Answer 1

避免除以 0 或取模

//while (n--) {
for ( ; n; n--)  {
  ...
  x = rand() % n;