嗨，我正在研究这段代码：
#include <signal.h>
#include <stdio.h>
#include <string.h>
#include <sys/time.h>
#include <unistd.h>
#include <stdlib.h>
#include <stdbool.h>
volatile sig_atomic_t print_flag = true;
static int count = 0;
void timer_handler (int signum)
{
printf ("timer expired %d times\n", ++count);
if(count>20) {
print_flag = false;
}
}
int main ()
{
struct sigaction sa;
struct itimerval timer;
/* Install timer_handler as the signal handler for SIGVTALRM. */
memset (&sa, 0, sizeof (sa));
sa.sa_handler = &timer_handler;
sigaction (SIGALRM, &sa, NULL);
/* Configure the timer to expire after 250 msec... */
timer.it_value.tv_sec = 0;
timer.it_value.tv_usec = 250000;
/* ... and every 250 msec after that. */
timer.it_interval.tv_sec = 0;
timer.it_interval.tv_usec = 250000;
/* Start a virtual timer. It counts down whenever this process is
executing. */
setitimer (ITIMER_REAL, &timer, NULL);
/* Do busy work. */
while (print_flag) {
sleep(1);
}
printf("job done bye bye\n");
exit(0);
}
有了这个设置，一切都很好，我得到了这个输出
...
timer expired 17 times
timer expired 18 times
timer expired 19 times
timer expired 20 times
timer expired 21 times
job done bye bye
如果我尝试更改代码注释timer.it_interval.tv_usec和 timer.it_interval.tv_usec设置两者 timer.it_value.tv_sec和timer.it_value.tv_sec等于，例如，3 它不会做他的工作。
但是，如果我维护这样的明确声明，tv_usec它会起作用：
timer.it_value.tv_sec = 3;
timer.it_value.tv_usec = 0;
timer.it_interval.tv_sec = 3;
timer.it_interval.tv_usec = 0;
为什么我tv_usec对这两个字段的显式声明有约束力？