因此,有Master和Slave Attiny2313。主机发送9位数据(第9位TXB8设置为1),但从机未检测到第9位(RXB8仍为0)。
我认为,如果主机中的TXB8位置1,从机上的RXB8位置是否应该自动设置? (在代码中,我检查UCSRB中的RXB8是否设置为1。
这不是问题所在)
void USART_Init(void)
{
/* Set baud rate */
UBRRH = (unsigned char)(BAUDRATE>>8);
UBRRL = (unsigned char)BAUDRATE;
/* Enable receiver and transmitter */
UCSRB = (1<<RXEN)|(1<<TXEN);
/* Set frame format: 9data, 1stop bit */
UCSRC = (7<<UCSZ0);
}
void USART_Transmit(unsigned int data)
{
/* Wait for empty transmit buffer */
while ( !( UCSRA & (1<<UDRE)) );
/* Copy 9th bit to TXB8 */
UCSRB &= ~(1<<TXB8);
if ( data & 0x0100 )
UCSRB |= (1<<TXB8);
/* Put data into buffer, sends the data */
UDR = data;
//Slave Receive Code
gned int USART_Receive( void )
{
unsigned char status, resh, resl;
/* Wait for data to be received */
while ( !(UCSRA & (1<<RXC)) );
/* Get status and 9th bit, then data from buffer */
status = UCSRA;
resh = UCSRB;
resl = UDR;
return resh; ///test
/* If error, return -1 */
if ( status & ((1<<FE)|(1<<DOR)|(1<<UPE)) )
return -1;
/* Filter the 9th bit, then return */
resh = (resh >> 1) & 0x01;
return ((resh << 8) | resl);
}
最佳答案
我发现了问题所在,USART的初始化如下:
/* Enable receiver and transmitter */
UCSRB = (1<<RXEN)|(1<<TXEN);
/* Set frame format: 9data, 1stop bit */
UCSRC = (7<<UCSZ0);
但是UCSZ2位在USCRB寄存器中,而不在UCSRC中,因此正确的代码将是:
UCSRB = (1<<RXEN)|(1<<TXEN)|(1<<UCSZ2);
UCSRC = (1<<UCSZ0)|(1<<UCSZ1);
关于c - AVR 9位Usart不起作用(MPCM),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/33431374/