workqueue的问题讨论

发布于 2022-10-20 11:01:51 浏览：401 订阅该版

```c
rt_inline rt_err_t _workqueue_work_completion(struct rt_workqueue *queue)
{
    rt_err_t result;

rt_enter_critical();
    while (1)
    {
        /* try to take condition semaphore */
        result = rt_sem_trytake(&(queue->sem));
        if (result == -RT_ETIMEOUT)
        {
            /* it's timeout, release this semaphore */
            rt_sem_release(&(queue->sem));
        }
        else if (result == RT_EOK)
        {
            /* keep the sem value = 0 */
            result = RT_EOK;
            break;
        }
        else
        {
            result = -RT_ERROR;
            break;
        }
    }
    rt_exit_critical();

return result;
}
```
这段代码`queue->sem`信号量的处理问题。
该函数进入时信号量值为0，退出时还是为0，那么该函数的意义何在那？
用到该信号量的地方只有在如下代码中：
```c
rt_err_t rt_workqueue_cancel_work_sync(struct rt_workqueue *queue, struct rt_work *work)
{
    RT_ASSERT(queue != RT_NULL);
    RT_ASSERT(work != RT_NULL);

if (queue->work_current == work) /* it's current work in the queue */
    {
        /* wait for work completion */
        rt_sem_take(&(queue->sem), RT_WAITING_FOREVER);
    }
    else
    {
        _workqueue_cancel_work(queue, work);
    }

return RT_EOK;
}
```
按照上面分析`rt_sem_take(&(queue->sem), RT_WAITING_FOREVER); `这个永远等不到信号量。