C 为什么timerfd定期Linux计时器比预期提前一点过期?
我使用的是Linux定期计时器,特别是,C 为什么timerfd定期Linux计时器比预期提前一点过期?,c,linux,timer,gettimeofday,C,Linux,Timer,Gettimeofday,我使用的是Linux定期计时器,特别是,timerfd,我将其设置为定期过期,例如每200毫秒一次 然而,我注意到计时器似乎有时会在我设置的超时时间之前一点过期 特别是,我使用以下C代码执行一个简单的测试: #include <stdlib.h> #include <stdio.h> #include <time.h> #include <poll.h> #include <unistd.h> #include <inttypes
timerfd
,我将其设置为定期过期,例如每200毫秒一次
然而,我注意到计时器似乎有时会在我设置的超时时间之前一点过期
特别是,我使用以下C代码执行一个简单的测试:
#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;
}
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()
估计的实时差异永远不会小于200ms(这也是因为用read()
清除事件所需的时间),但也有一些值略小于200ms,如199.942000ms
你知道我为什么要观察这种行为吗
这可能是因为我使用的是gettimeofday()
,有时,tv\u prev
会延迟一点(由于调用read()
或gettimeofday()
本身时会有不同的延迟),而tv\u curr
在下一次迭代中不会延迟,导致估计的时间小于200ms,而计时器实际上是精确的,每200毫秒过期一次
提前非常感谢。这与进程调度有关。计时器确实非常精确,每200毫秒发出一次超时信号,但在实际获得控制权之前,程序不会注册该信号。这意味着您从
gettimeofday()
调用中获得的时间可以显示将来的某个时刻。当你减去这些延迟值,你可以得到大于或小于200毫秒的结果
如何估计计时器的实际信号与调用gettimeofday()
之间的时间?它与进程调度的时间量有关。此量程有一些由中的RR_时间片设置的默认值。您可以通过以下方式在系统上进行检查:
#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,然后才能获得控件并读取当前时间。在我的系统上,它可能导致产生的延迟与预期的200毫秒相差约±4毫秒。
在执行了近7000次迭代之后,我得到了注册等待时间的以下分布:
如您所见,大多数时间都在预期200 ms左右的±2 ms间隔内。所有迭代中的最小和最大时间分别为189.992 ms和210.227 ms:
~$ sort times.txt | head
189.992000
190.092000
190.720000
194.402000
195.250000
195.746000
195.847000
195.964000
196.256000
196.420000
~$ sort times.txt | tail
203.746000
203.824000
203.900000
204.026000
204.273000
205.625000
205.634000
208.974000
210.202000
210.227000
~$
大于4毫秒的偏差是由于程序需要等待几个量程而不是一个量程的罕见情况造成的。我认为您的猜测是正确的。请注意,timerfd不会等待您在下一个计时器启动之前读取它。Linux不是实时操作系统。预计时间会是“差不多”的——200多岁的女士。在特殊情况下,我不会感到惊讶,因为它会被更多的人淘汰。这比Windows好多了。
Scheduler quantum is 4.000000 ms
~$ sort times.txt | head
189.992000
190.092000
190.720000
194.402000
195.250000
195.746000
195.847000
195.964000
196.256000
196.420000
~$ sort times.txt | tail
203.746000
203.824000
203.900000
204.026000
204.273000
205.625000
205.634000
208.974000
210.202000
210.227000
~$