关于rt_thread_timer_entry函数中next_tick的问题

发布于 2021-08-26 11:03:23 浏览：979 订阅该版

```c
/* system timer thread entry */
static void rt_thread_timer_entry(void *parameter)
{
    rt_tick_t next_timeout;

while (1)
    {
        /* get the next timeout tick */
        next_timeout = rt_timer_list_next_timeout(rt_soft_timer_list);
        if (next_timeout == RT_TICK_MAX)
        {
            /* no software timer exist, suspend self. */
            rt_thread_suspend(rt_thread_self());
            rt_schedule();
        }
        else
        {
            rt_tick_t current_tick;

/* get current tick */
            current_tick = rt_tick_get();

if ((next_timeout - current_tick) < RT_TICK_MAX / 2)
            {
                /* get the delta timeout tick */
                next_timeout = next_timeout - current_tick;
                rt_thread_delay(next_timeout);
            }
        }

/* check software timer */
        rt_soft_timer_check();
    }
}
```
变量next_timeout是定时器下一次溢出的时间，如果我20ms溢出一次，然后当前时间是1000ms，这样做减法，会溢出吧。
下面还有如果相减的结果小于多少，这样减完是负数啊