C ++和汇编程序：测量mips

Question

我需要在C ++中使用内联汇编程序每秒测量数百万次操作，我是两者中的新手，所以我的程序有一些不一致的输出，但我不知道，输出应该关闭的是什么数字来，所以我无法真正比​​较它们。

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

using namespace std;
int main()
{   clock_t t1, t2, t3;
    long n = 500000000;
    long i = 0;

    t1 = clock();
    for(i = 0; i < n; i++){
        __asm__ volatile (
        ".intel_syntax noprefix\n\t"
        "mov ax, bx\n\t"
        );
    }
    t2 = clock();

    for(i = 0; i < n; i++){
        __asm__ volatile (
        ".intel_syntax noprefix\n\t"
        "mov ax, bx\n\t"
        "mov bx, ax\n\t"
        );
    }
    t3 = clock();

    int d1 = t2 - t1;
    int d2 = t3 - t2;
    double operation = (d2 - d1) / (double) n;

    cout <<"n + loop:    " << d1 << " ms" << endl;
    cout <<"2n + loop:   " << d2 << " ms" << endl;
    cout << "n operations:  " << d2 - d1 << " ms" << endl;
    cout << "1 operation:  " << operation << " ms" << endl;
    cout << "mops:  "<< 1000 / (operation * 1024 * 1024) <<endl;
    return (0);
}

输出就像：

n + loop:    1590078 ms
2n + loop:   1598367 ms
n operations:  8289 ms
1 operation:  1.6578e-05 ms
mops:  57.5265

或

n + loop:    1593836 ms
2n + loop:   1593969 ms
n operations:  133 ms
1 operation:  2.66e-07 ms
mops:  3585.24

或

n + loop:    1582123 ms
2n + loop:   1592557 ms
n operations:  10434 ms
1 operation:  2.0868e-05 ms
mops:  45.7003

所以我的问题基本上是：它是一种适当的方法来测量，甚至大致是mips？