STM32串口中斷接收一個完整的數(shù)據(jù)幀

（1）大端發(fā)送；
（2）上位機發(fā)送一幀數(shù)據(jù)的時間間隔不能大于主循環(huán)周期；

（3）數(shù)據(jù)幀滿足下面格式：

void USART6_Init (void) {GPIO_InitTypeDef	GPIO_InitStructure;USART_InitTypeDef	USART_InitStructure;NVIC_InitTypeDef	NVIC_InitStructure;RCC_APB2PeriphClockCmd(RCC_APB2Periph_USART6,ENABLE);RCC_AHB1PeriphClockCmd(RCC_AHB1Periph_GPIOC,ENABLE); //修改GPIO_InitStructure.GPIO_Pin	= GPIO_Pin_6|GPIO_Pin_7;//修改GPIO_InitStructure.GPIO_Mode	= GPIO_Mode_AF;GPIO_InitStructure.GPIO_OType	= GPIO_OType_PP;GPIO_InitStructure.GPIO_PuPd	= GPIO_PuPd_NOPULL;GPIO_InitStructure.GPIO_Speed	= GPIO_Speed_50MHz;GPIO_Init(GPIOC,&GPIO_InitStructure);//修改GPIO_PinAFConfig(GPIOC,GPIO_PinSource6,GPIO_AF_USART6);//修改GPIO_PinAFConfig(GPIOC,GPIO_PinSource7,GPIO_AF_USART6);	//修改NVIC_InitStructure.NVIC_IRQChannel = USART6_IRQn;NVIC_InitStructure.NVIC_IRQChannelPreemptionPriority = 2;NVIC_InitStructure.NVIC_IRQChannelSubPriority = 2;NVIC_InitStructure.NVIC_IRQChannelCmd = ENABLE;NVIC_Init(&NVIC_InitStructure);USART_InitStructure.USART_BaudRate = 115200;USART_InitStructure.USART_HardwareFlowControl = USART_HardwareFlowControl_None;USART_InitStructure.USART_Mode = USART_Mode_Tx|USART_Mode_Rx;USART_InitStructure.USART_Parity = USART_Parity_No;USART_InitStructure.USART_StopBits = USART_StopBits_1;USART_InitStructure.USART_WordLength = USART_WordLength_8b;USART_Init(USART6,&USART_InitStructure);USART_ITConfig(USART6,USART_IT_RXNE,ENABLE);//打開接收中斷USART_Cmd(USART6,ENABLE);}void USART6_IRQHandler(){unsigned char rCh;static char rCnt = 0;if(USART_GetITStatus(USART6,USART_IT_RXNE) != RESET){rCh = USART_ReceiveData(USART6);COM6_RecvBuf[rCnt] = rCh;if(rCnt == 0)     //幀頭0xAA    {rCnt = (0xAA != rCh)?0:rCnt+1;}else if(rCnt == 1) //幀頭0x55  {rCnt = (0x55 != rCh)?0:rCnt+1;}else if(rCnt == 2) //類型type{//這里可以根據(jù)類型的范圍進行如上的處理rCnt++;}else if(rCnt == 3) //長度len{rCnt++;}			else if(rCnt > 3) //值value{rCnt++;if(rCnt == 6+COM6_RecvBuf[3]){             rCnt = 0;memcpy(COM6_RecvBufBck,COM6_RecvBuf,RECV_BUF_SZ);//緩沖COM6_RecvFin = 1;  //通知主循環(huán)處理}}}}int  main(void){int i;//代碼段1while(1)//該循環(huán)不能太慢，否則數(shù)據(jù)緩沖區(qū)會被部分修改{//代碼段2if(COM6_RecvFin == 1){COM6_RecvFin = 0;CMD_Analysis();//分析接收到的這幀數(shù)據(jù)}//代碼段3}return  0;}//在以后再仔細分析數(shù)據(jù)接收較快而處理較慢的問題吧，本課題主要討論的是如何完整的接收一個數(shù)據(jù)幀，在數(shù)據(jù)源正確的情況下不丟幀