問題描述：

最近項(xiàng)目里使用了兩路串口透傳數(shù)據(jù)，串口3中斷接收數(shù)據(jù)，組包后通過串口1阻塞發(fā)送出去。組完包的數(shù)據(jù)近800個(gè)字符，哼唧狗使用HAL_UART_Transmit()函數(shù)進(jìn)行發(fā)送，理論和實(shí)際驗(yàn)證發(fā)送完一包數(shù)據(jù)115200波特率下需要耗時(shí)近70ms。

哼唧狗一直有個(gè)疑問，在串口1發(fā)送的70ms內(nèi)，串口3的中斷接收會(huì)不會(huì)受影響呢？

這個(gè)疑問困擾了自己好幾天了，由于自己對(duì)串口理解不深，今晚實(shí)在忍不住就去芯片之家群里問大牛們?nèi)チ恕?/p>

得到的一個(gè)大神回復(fù)是這樣的：

** 下面是測試過程分享。**

串口1阻塞發(fā)送800個(gè)字符，然后串口3中斷接收，每次接收到一個(gè)字符變量加1，并在串口中斷回調(diào)函數(shù)內(nèi)打印該變量值。通過串口助手給串口3發(fā)送數(shù)據(jù)，每次發(fā)送一個(gè)字符。看發(fā)送過程中是否會(huì)受影響。

main函數(shù)中：

while (1)
    {
        //800個(gè)字符 理論串口發(fā)送70ms
        sprintf(buf,"%s","AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAn");      
        while(1)
        {
            HAL_UART_Transmit(&huart1, (uint8_t *)buf, strlen(buf),1000);
            //加一個(gè)指示燈翻轉(zhuǎn)
            HAL_GPIO_TogglePin(LED0_GPIO_Port,LED0_Pin);
        }
    }

串口1中斷回調(diào)函數(shù)：

int conut = 0;
void HAL_UART_RxCpltCallback(UART_HandleTypeDef *huart)
{
    if(huart == &huart3)
    {
        conut = conut + 1;
        printf("conut = %d n ",conut);
        HAL_UART_Receive_IT(&huart3, (uint8_t *)Rx2478Tmp, RxDataTmp);
    }
}

串口助手模擬發(fā)送：

測試結(jié)果：

不受影響。