我正在使用Linux定期计时器,尤其是timerfd
,我将其设置为定期过期,例如每200毫秒一次。
但是,我发现计时器似乎有时会在我设置的超时时间之前过期。
特别是,我正在使用以下C代码执行简单的测试:
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <poll.h>
#include <unistd.h>
#include <inttypes.h>
#include <sys/timerfd.h>
#include <sys/time.h>
#define NO_FLAGS_TIMER 0
#define NUM_TESTS 10
// Function to perform the difference between two struct timeval.
// The operation which is performed is out = out - in
static inline int timevalSub(struct timeval *in, struct timeval *out) {
time_t original_out_tv_sec=out->tv_sec;
if (out->tv_usec < in->tv_usec) {
int nsec = (in->tv_usec - out->tv_usec) / 1000000 + 1;
in->tv_usec -= 1000000 * nsec;
in->tv_sec += nsec;
}
if (out->tv_usec - in->tv_usec > 1000000) {
int nsec = (out->tv_usec - in->tv_usec) / 1000000;
in->tv_usec += 1000000 * nsec;
in->tv_sec -= nsec;
}
out->tv_sec-=in->tv_sec;
out->tv_usec-=in->tv_usec;
// '1' is returned when the result is negative
return original_out_tv_sec < in->tv_sec;
}
// Function to create a timerfd and set it with a periodic timeout of 'time_ms', in milliseconds
int timerCreateAndSet(struct pollfd *timerMon,int *clockFd,uint64_t time_ms) {
struct itimerspec new_value;
time_t sec;
long nanosec;
// Create monotonic (increasing) timer
*clockFd=timerfd_create(CLOCK_MONOTONIC,NO_FLAGS_TIMER);
if(*clockFd==-1) {
return -1;
}
// Convert time, in ms, to seconds and nanoseconds
sec=(time_t) ((time_ms)/1000);
nanosec=1000000*time_ms-sec*1000000000;
new_value.it_value.tv_nsec=nanosec;
new_value.it_value.tv_sec=sec;
new_value.it_interval.tv_nsec=nanosec;
new_value.it_interval.tv_sec=sec;
// Fill pollfd structure
timerMon->fd=*clockFd;
timerMon->revents=0;
timerMon->events=POLLIN;
// Start timer
if(timerfd_settime(*clockFd,NO_FLAGS_TIMER,&new_value,NULL)==-1) {
close(*clockFd);
return -2;
}
return 0;
}
int main(void) {
struct timeval tv,tv_prev,tv_curr;
int clockFd;
struct pollfd timerMon;
unsigned long long junk;
gettimeofday(&tv,NULL);
timerCreateAndSet(&timerMon,&clockFd,200); // 200 ms periodic expiration time
tv_prev=tv;
for(int a=0;a<NUM_TESTS;a++) {
// No error check on poll() just for the sake of brevity...
// The final code should contain a check on the return value of poll()
poll(&timerMon,1,-1);
(void) read(clockFd,&junk,sizeof(junk));
gettimeofday(&tv,NULL);
tv_curr=tv;
if(timevalSub(&tv_prev,&tv_curr)) {
fprintf(stdout,"Error! Negative timestamps. The test will be interrupted now.\n");
break;
}
printf("Iteration: %d - curr. timestamp: %lu.%lu - elapsed after %f ms - real est. delta_t %f ms\n",a,tv.tv_sec,tv.tv_usec,200.0,
(tv_curr.tv_sec*1000000+tv_curr.tv_usec)/1000.0);
tv_prev=tv;
}
return 0;
}
用gcc编译后:
gcc -o timertest_stackoverflow timertest_stackoverflow.c
我得到以下输出:
Iteration: 0 - curr. timestamp: 1583491102.833748 - elapsed after 200.000000 ms - real est. delta_t 200.112000 ms
Iteration: 1 - curr. timestamp: 1583491103.33690 - elapsed after 200.000000 ms - real est. delta_t 199.942000 ms
Iteration: 2 - curr. timestamp: 1583491103.233687 - elapsed after 200.000000 ms - real est. delta_t 199.997000 ms
Iteration: 3 - curr. timestamp: 1583491103.433737 - elapsed after 200.000000 ms - real est. delta_t 200.050000 ms
Iteration: 4 - curr. timestamp: 1583491103.633737 - elapsed after 200.000000 ms - real est. delta_t 200.000000 ms
Iteration: 5 - curr. timestamp: 1583491103.833701 - elapsed after 200.000000 ms - real est. delta_t 199.964000 ms
Iteration: 6 - curr. timestamp: 1583491104.33686 - elapsed after 200.000000 ms - real est. delta_t 199.985000 ms
Iteration: 7 - curr. timestamp: 1583491104.233745 - elapsed after 200.000000 ms - real est. delta_t 200.059000 ms
Iteration: 8 - curr. timestamp: 1583491104.433737 - elapsed after 200.000000 ms - real est. delta_t 199.992000 ms
Iteration: 9 - curr. timestamp: 1583491104.633736 - elapsed after 200.000000 ms - real est. delta_t 199.999000 ms
我希望用gettimeofday()
估算的实时差异永远不会小于200毫秒(也是由于用read()
清除事件所需的时间),但是也有一些差异小于200毫秒的值,例如199.942000 ms
。
您知道我为什么要观察这种行为吗?
这可能是由于我使用gettimeofday()
,有时tv_prev
待用较晚(由于调用read()
或{{ 1}}本身)和gettimeofday()
在下一次迭代中不会导致估计的时间少于200毫秒,而计时器实际上精确到每200毫秒过期一次?
非常感谢您。
答案 0 :(得分:0)
这与过程计划有关。计时器确实非常精确,并且每200毫秒发出一次超时信号,但是您的程序直到实际获得控制权后才会注册该信号。这意味着您从#include <Eigen/Core>
int main()
{
Eigen::ArrayXd a(10);
a.setRandom();
return (a != 0.0).any();
}
通话中获得的时间可以显示将来的某个时刻。当减去此类延迟值时,可以获得大于或小于200毫秒的结果。
如何估计计时器的实际信号与您致电gettimeofday()
之间的时间?它与时间的过程调度量有关。此量子具有include/linux/sched/rt.h中RR_TIMESLICE设置的某些默认值。您可以像这样在系统上对其进行检查:
gettimeofday()
我的系统上的输出:
#include <sched.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdio.h>
int main(void) {
struct timespec tp;
if (sched_rr_get_interval(getpid(), &tp)) {
perror("Cannot get scheduler quantum");
} else {
printf("Scheduler quantum is %f ms\n", (tp.tv_sec * 1e9 + tp.tv_nsec) / 1e6);
}
}
因此,您可能需要等待另一个进程的调度程序完成才能获得控件,然后才能读取当前时间。在我的系统上,这可能导致所产生的延迟与预期的200 ms发生大约±4 ms的偏差。 经过大约7000次迭代,我得到了注册等待时间的以下分布:
如您所见,大多数情况下,预期的200 ms间隔为±2 ms。所有迭代中的最小和最大时间分别为189.992 ms和210.227 ms:
Scheduler quantum is 4.000000 ms
大于4毫秒的偏差是由罕见情况引起的,这种情况是程序需要等待几个量子点而不仅仅是一个量子点。