关于超时退出循环的一个疑问

发布于 2019-11-19 18:51:41 浏览：1332 订阅该版

这是我的一个uart串口通信实验，是利于信号来实现的，我的想法是如果超过100ms 串口没有接收到数据的话就退出循环，这是我的源代码，但是没有实现我想要的效果，请大神看一下``` 
 sem = rt_sem_create(gst_juart.uart_param[1].sem_name, 1, RT_IPC_FLAG_FIFO);
 if (RT_NULL == sem)
 {
 rt_kprintf("JUSONBP_UART: fail to create rx_sem ");
 return;
 }
 gst_juart.uart_param[1].rx_sem = sem;
 ```

``` while (1) {
 err = rt_sem_take(sem, 100);
 if (-RT_ETIMEOUT == err)
					{
							 
//					sem = rt_sem_create(gst_juart.uart_param[1].sem_name, 1, RT_IPC_FLAG_FIFO);
//					if (RT_NULL == sem)
//					{
//						rt_kprintf("JUSONBP_UART: fail to create rx_sem ");
// // return;
//					}
//						gst_juart.uart_param[1].rx_sem = sem;
// 
							 resp_len = snprintf(resp_buf ,resp_buf_len ,"<CMD_ERR>");
 
 bpl_kprintf(BP_DEBUG_LEVEL_UART, "UART: sem is timeout ");
 break;
 }
 
 memset(read_buf, 0, sizeof(read_buf));
 rt_device_read(dev, 0, read_buf, sizeof(read_buf)); 
 if ('>' == read_buf[0]) {
						 resp_len += snprintf(resp_buf + resp_len, resp_buf_len - resp_len,
 "%c", '>');
 bpl_kprintf(BP_DEBUG_LEVEL_UART, "UART: read finished. ");
 break;
 } else {
 bpl_kprintf(BP_DEBUG_LEVEL_UART, "UART: read: %s ", read_buf);
 }
 
 resp_len += snprintf(resp_buf + resp_len, resp_buf_len - resp_len,
 "%s", read_buf);
//				rt_kprintf("%d ",i++);
//	rt_thread_delay(1);
	} ```