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uint16_t USART_ReceiveData(USART_TypeDef* USARTx)0 P# z9 {% L8 b# w3 B8 G- p
{
0 Y _ w. }. S7 G7 e) g$ C: l2 i /* Check the parameters */; F* r, B- x/ h, {. U
assert_param(IS_USART_ALL_PERIPH(USARTx));
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' d _( L3 X. E- v* J# M /* Receive Data */: m" q! v+ W; }- l
return (uint16_t)(USARTx->DR & (uint16_t)0x01FF);! T5 S( t N( o, A
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* c# b+ \! m9 h' s' \; Z0 d- \0 T为什么要返回16位,串口一次接8位呀。用的时候也是用8位,谁帮忙解释一下
7 z& V9 S% b, s" O* muint8_t uart_buff[UART_BUFF_SIZE];
5 K, V, u/ m0 i o: duart_buff[uart_p] = USART_ReceiveData(BLT_USARTx);/ U( ?, M" G( n) K- `$ U0 D
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