CPU计算性能测试程序

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;
	}
}