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问 我在xt2端接3。6864MHz晶振，想用smclk做uart的时钟，速率设为9600，
但是不好用，请帮忙，程序如下：

WDTCTL=WDTPW+WDTHOLD;
BCSCTL1 &= ~XT2OFF; // XT2on
BCSCTL1 |= SELM1+SELS;; // ACLK = LFXT1 = HF XTAL
UCTL1 = CHAR; // 8-bit character
UTCTL1 = SSEL1; // UCLK = ACLK
UBR01 = 0x80; //3686400/9600 - 0x180
UBR11 = 0x01; //
UMCTL1 = 0x00; // Modulation
ME2 |= UTXE1 + URXE1; // Enable USART0 TXD/RXD

P3SEL |= 0xc0;
P3DIR |= 0x40;

IE2 |= URXIE1; 答 1: BCSCTL1 |= SELM1+SELS;; // ACLK = LFXT1 = HF XTAL

应该是BCSCTL2吧。 答 2: BCSCTL2.
SELS位为1直接选择XT2，MCLK要选择XT2需加上
do
{
IFG1 &= ~OFIFG; // Clear OSCFault flag
for (i = 0xFF; i > 0; i--); // Time for flag to set
}
while ((IFG1 & OFIFG) != 0); // OSCFault flag still set? 答 3: 笔误了