CPU计算性能测试程序

来源：互联网发布：子见夫子乎编辑：程序博客网时间：2024/04/29 12:28

计算下面一系列的数学计算在1s中能执行过少次。

#include "stdio.h"#include "time.h"int i, j, l, k, m, jj;jj = 2342;k = 31455;l = 16452;m = 9823;i = 1000000;void main() {int warp_count = 0;int max_warp = 1000;long int count = 0;time_t b_second,l_second;time_t rawtime;struct tm * timeinfo;while(1){b_second = time(NULL);l_second = b_second+1;while ((b_second=time(NULL))<l_second) {m = m ^ l;k = (k / m * jj) % i;l = j * m * k;i = (j * k) ^ m;k = (k / m * jj) % i;m = m ^ l;m = m ^ l;i = (j * k) ^ m;k = (k / m * jj) % i;m = i * i * i * i * i * i * i; // m=k*l*jj*l;m = m ^ l;k = (k / m * jj) % i;l = j * m * k;i = (j * k) ^ m;l = (k / m * jj) % i;m = m ^ l;m = m ^ l;i = (j * k) ^ m;k = (k / m * jj) % i;m = k * k * k * k * k - m / i;count++;}time(&rawtime);timeinfo = localtime (&rawtime);printf("Time: %s ", asctime (timeinfo));printf("%ld\n",count);count=0;warp_count++;if(warp_count==max_warp)break;}}

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