如何使用套接字编程和 UDP 套接字以 512 纳秒的延迟发送数据包

Question

纳米睡眠的问题在于它包含相当大的开销。用户往往期望延迟精确到非常低的值，这对于 CPU 产生睡眠例程来说并不实际。期望如下：

actual delay = nanosleep(requested delay)

然而，现实中用户得到的是这样的：

actual delay = nanosleep(requested delay) + the fixed overheads.

在我的测试计算机上，强制 CPU 亲和性（以消除调度程序的影响）并将 CPU 频率缩放调节器设置为性能，固定开销为 52 微秒。

解决您问题的一个可能方法是用延迟循环替换对 nanosleep 的调用。这样 CPU 就不会让步，进入空闲状态，也不会产生操作系统调用开销。当然，延迟循环可能难以校准，无法真正获得所需的延迟。我制作了两个程序，一个使用 nanosleep，一个使用延迟循环进行演示。为了使其他开销可以忽略不计，每次执行时间和打印类型调用时，睡眠和延迟循环都会运行很多次：

作为参考，所使用的程序：

纳米睡眠版本：

/*****************************************************************************
*
* test_slp2.c 2019.09.18 Smythies
*       Create data to determine the lower limit of nanosleep.
*
* test_slp.cpp 2017.11.25 Smythies
*       Of course, none of this stuff works the way it used to.
*
* test_slp.cpp 2012.01.24 Smythies
*          I need to be able to yeild (sleep), but for less than a second.
*          Experiment with nanosleep and usleep functions.
*
*****************************************************************************/

// prevent warning about nanosleep
#define _POSIX_C_SOURCE 199309L

#include <sys/time.h>
#include <sys/types.h>
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>

#define CR 13

unsigned long long stamp(void){
   struct timeval tv;

   gettimeofday(&tv, NULL);

   return (unsigned long long)tv.tv_sec * 1000000 + tv.tv_usec;
} /* endprocedure */

int main(){
   unsigned long long start, now;
   long i, j, k;
   int ns;
   struct timespec time;

   time.tv_sec = 0;
   start = stamp();
   for(ns = 500; ns < 100000; ns = ns + 500){
      for(j = 0; j < 100000; j++){
         time.tv_nsec = ns;
         nanosleep(&time, &time);
      } /* endfor */
      now = stamp();
      printf("%d %llu\n", ns, (now - start));
      start = now;
   } /* endfor */
   return(0);
}

而延迟循环版本，包括粗略的校准，如图所示，效果并不好：

/*****************************************************************************
*
* test_slp3.c 2019.09.18 Smythies
*       Now, do the same as test_slp2, but use a waste time loop instead of
        nanosleep.
*
* test_slp2.c 2019.09.18 Smythies
*       Create data to determine the lower limit of nanosleep.
*
* test_slp.cpp 2017.11.25 Smythies
*       Of course, none of this stuff works the way it used to.
*
* test_slp.cpp 2012.01.24 Smythies
*          I need to be able to yeild (sleep), but for less than a second.
*          Experiment with nanosleep and usleep functions.
*
*****************************************************************************/

// prevent warning about nanosleep
#define _POSIX_C_SOURCE 199309L

#include <sys/time.h>
#include <sys/types.h>
#include <time.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <errno.h>

/* for my test computer */
#define CALIBRATION 26

unsigned long long stamp(void){
   struct timeval tv;

   gettimeofday(&tv, NULL);

   return (unsigned long long)tv.tv_sec * 1000000 + tv.tv_usec;
} /* endprocedure */

int main(){
   unsigned long long start, now;
   long i, j, k, m, loops;
   int ns;
   struct timespec time;

   time.tv_sec = 0;
   start = stamp();
   for(ns = 500; ns <= 10000; ns = ns + 100){
      loops = ns * CALIBRATION / 50;  /* will have rounding issues */
      for(m = 0; m < 10000000; m++){ /* this is just to slow things down, so we can test */
         for(j = 0; j < loops; j++){
            k = j;  /* make sure any compile optimizer doesn't take out the loop */
         } /* endfor */
      } /* endfor */
      now = stamp();
      printf("%d %llu\n", ns, (now - start));
      start = now;
   } /* endfor */
   return(0);
}

预期问题：为什么延迟循环测试中没有 gettimeofday 抖动，而您（我）声称这是 nanosleep 版本中抖动的原因？答案：可能是因为延迟循环方法每个样本的运行时间比 nanosleep 版本长得多。或者，我错了。

这些测试强制使用 CPU 亲和性。例如：

time taskset -c 3 ./test_slp3

我对整个测试进行了计时，只是为了进行合理性检查，因为各个时间的总和应该等于总时间。他们做到了。

CPU 频率调节调节器设置为性能：

$ cat /sys/devices/system/cpu/cpu*/cpufreq/scaling_governor
performance
performance
performance
performance
performance
performance
performance
performance

Answer 1