Question

我对libev的ev_timer有疑问，请看下面的代码：

static void timeout_cb(struct ev_loop *loop, ev_timer *timer, int revent) {
    printf("got an timeout event, current time %s\n", get_current_time());
}

int main(int argc, char *argv[]) {
    struct ev_loop *loop = ev_loop_new(0);
    ev_timer_init(&timer, timeout_cb, 5.0, 0.0);
    ev_timer_start(loop, &timer);
    ev_run(loop, EVRUN_NOWAIT);  // callback should not be called
    ev_timer_stop(loop, &timer); // stop the timer
    sleep(5);  // sleep 5 seconds, 5 is also the timer's timeout value
    // restart timer
    ev_timer_init(&timer, timeout_cb, 5.0, 0.0);
    ev_timer_start(loop, &timer);
    printf("timer start at: %s\n", get_current_time());
    printf("timer remaining: %f\n", ev_timer_remaining(loop, &timer));
    ev_run(loop, EVRUN_NOWAIT);
    return 0;
}

输出结果为：

timer start at: 14:53:49:137
timer remaining: 5.0000
got an timeout event, current time 14:53:49:137

这很奇怪，因为在重新启动计时器后，计时器立即触发，但应该在5秒之后。我发现原因是睡眠（5）。我将其更改为sleep（4），然后不会调用计时器回调。我对liebev的计时器功能感到困惑。我对计时器有误解吗？如何在定时器重启后超时后调用定时器回调？

Answer 1

您应该更改代码

ev_timer_init(&timer, timeout_cb, 5.0, 0.0);

到

ev_timer_init(&timer, timeout_cb, ev_time() - ev_now() + 5.0, 0.0);

因为libev会因性能原因缓存时间戳，并且只在每个事件循环之前或之后更新时间戳。在您休眠五秒后，当前实时将为T + 5，即ev_time，但libev的当前时间戳仍为T，即ev_now。那么你就像这样启动一个计时器

ev_timer_init(&timer, timeout_cb, 5.0, 0.0);

libev会将当前时间戳加上5的定时器触发时间设置为T + 5，也是当前实时，因此会立即触发。但是当你像这样启动计时器时

ev_timer_init(&timer, timeout_cb, ev_time() - ev_now() + 5.0, 0.0);

libev将在当前时间戳脉冲ev_time() - ev_now() + 5.0设置定时器的触发时间，即T + T + 5 - T + 5 = T + 5 + 5 = real time + 5。所以它会在五秒钟后触发。

Answer 2

我并不是自称是libev的专家，但我发现使用ev_suspend（循环）和ev_resume（循环）会阻止你看到的行为。从手册页：

ev_suspend (loop)
ev_resume (loop)
    These two functions suspend and resume an event loop, for use when
    the loop is not used for a while and timeouts should not be
    processed.

E.g：

ev_run(loop, EVRUN_NOWAIT);  // callback should not be called
ev_timer_stop(loop, &timer); // stop the timer
ev_suspend(loop); // suspend timer processing
sleep(5);  // sleep 5 seconds, 5 is also the timer's timeout value
ev_resume(loop); // resume timer processing
// restart timer
ev_timer_init(&timer, timeout_cb, 5.0, 0.0);
ev_timer_start(loop, &timer);

此外，我发现另一种可能的解决方案是使用ev_timer_again（）而不是ev_timer_start（），这也会阻止您报告的行为。

E.g：

ev_timer_stop(loop, &timer); // stop the timer
sleep(5);  // sleep 5 seconds, 5 is also the timer's timeout value
// restart timer
ev_timer_again(loop, &timer);

ev_timer回调的调用时间早于它应该调用

2 个答案: