在Ubuntu/VMware上测量C中的短时间间隔

Question

我试图以几毫秒的顺序测量短时间间隔.这被证明是一项非常重要的任务,因为无处不在的time()功能在整个秒精度上起作用.经过一些研究,我得出了以下代码示例中的四种方法:

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

int main(int argc, char *argv[])
{
    clock_t t0, t1;
    double dt0;

    struct timespec t2, t3;
    double dt1;

    struct timespec t4, t5;
    double dt2;

    struct timeval t6, t7;
    double dt3;

    volatile long long i;

    t2.tv_sec = 0;
    t2.tv_nsec = 0;
    clock_settime(CLOCK_MONOTONIC, &t2);
    clock_settime(CLOCK_REALTIME,  &t2);

    t0 = clock();
    clock_gettime(CLOCK_REALTIME,  &t2);
    clock_gettime(CLOCK_MONOTONIC, &t4);
    gettimeofday(&t6, NULL);

    getchar();
    for (i=0; i<1e9; i++) {};

    gettimeofday(&t7, NULL);
    clock_gettime(CLOCK_MONOTONIC, &t5);
    clock_gettime(CLOCK_REALTIME,  &t3);
    t1 = clock();

    dt0 = (double) (t1 - t0) / CLOCKS_PER_SEC;
    dt1 = (double) (t3.tv_nsec - t2.tv_nsec) / 1e9;
    dt2 = (double) (t5.tv_nsec - t4.tv_nsec) / 1e9;
    dt3 = (double) (t7.tv_usec - t6.tv_usec) / 1e6;

    printf("1. clock():          [0]=%10.0f, [1]=%10.0f, [1-0]=%10.6f sec\n", (double) t0, (double) t1, dt0);
    printf("2. clock_gettime(R): [2]=%10ld, [3]=%10ld, [3-2]=%10f sec\n", (long) t2.tv_nsec, (long) t3.tv_nsec, dt1);
    printf("3. clock_gettime(M): [2]=%10ld, [3]=%10ld, [3-2]=%10f sec\n", (long) t4.tv_nsec, (long) t5.tv_nsec, dt2);
    printf("4. gettimeofday():   [4]=%10ld, [5]=%10ld, [5-4]=%10f sec\n", (long) t6.tv_usec, (long) t7.tv_usec, dt3);

    return 0;
}

然后,我编译并运行它:

gcc -lrt -o timer.e timer.c; time ./timer.e

我用getchar()电话来介绍几秒钟的延迟,看之间的差异real和user时间报告,其结果是:

1. clock():          [0]=         0, [1]=   3280000, [1-0]=  3.280000 sec
2. clock_gettime(R): [2]= 823922476, [3]= 478650549, [3-2]= -0.345272 sec
3. clock_gettime(M): [2]= 671137949, [3]= 325864897, [3-2]= -0.345273 sec
4. gettimeofday():   [4]=    823924, [5]=    478648, [5-4]= -0.345276 sec

real    0m6.659s
user    0m3.280s
sys     0m0.010s

如您所见,唯一有意义的结果是方法1与命令user报告的时间之间的相关性time.

这带来了几个问题:1.为什么结果2-4毫无意义？2.如何衡量运行程序所花费的实际时间,例如real数字time报告？

在基于AMD的惠普笔记本电脑上,我的环境是基于Windows 7的VMware上的Ubuntu 10.04 LTS 64位.

Answer 1

你正走在正确的轨道上clock_gettime().

但是,您应该摆脱这些clock_settime()调用:只需clock_gettime()在代码块之前和之后调用,然后查看差异.

此外,计算的差别时,你应该采取两个tv_sec和tv_nsec考虑; 您当前的代码只是查看纳秒组件并忽略整秒.

在我的Ubuntu系统上,以下代码可以非常准确地测量执行循环所需的时间:

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

int main(int argc, char *argv[])
{
    struct timespec t2, t3;
    double dt1;

    volatile long long i;

    clock_gettime(CLOCK_MONOTONIC,  &t2);
    for (i=0; i<1e9; i++) {};
    clock_gettime(CLOCK_MONOTONIC,  &t3);
    //time in seconds
    dt1 = (t3.tv_sec - t2.tv_sec) + (double) (t3.tv_nsec - t2.tv_nsec) * 1e-9;
    printf("%f\n", dt1);

    return 0;
}