刚接触msp430 2132程序如下就是 TI的例程改了一下用IAR软仿真：
//******************************************************************************
// MSP430F21x2 Demo - USCI_A0, Ultra-Low Pwr UART 9600 String, 32kHz ACLK
//
// Description: This program demonstrates a full-duplex 9600-baud UART using
// USCI_A0 and a 32kHz crystal. The program will wait in LPM3, and will
// respond to a received 'u' character using 8N1 protocol. The response will
// be the string 'Hello World'.
// ACLK = BRCLK = LFXT1 = 32768Hz, MCLK = SMCLK = DCO ~1.2MHz
// Baud rate divider with 32768Hz XTAL @9600 = 32768Hz/9600 = 3.41
// //* An external watch crystal is required on XIN XOUT for ACLK *//
//
// MSP430F21x2
// -----------------
// /|\| XIN|-
// | | | 32kHz
// --|RST XOUT|-
// | |
// | P3.4/UCA0TXD|------------>
// | | 9600 - 8N1
// | P3.5/UCA0RXD|<------------
//
// A. Dannenberg
// Texas Instruments Inc.
// October 2006
// Built with CCE Version: 3.2.0 and IAR Embedded Workbench Version: 3.41A
//******************************************************************************
#include "msp430x21x2.h"
const char string1[] = { "Hello World\r\n" };
unsigned int i;
void main(void)
{
WDTCTL = WDTPW + WDTHOLD; // Stop WDT
P1DIR = 0xFF; // All P1.x outputs
P1OUT = 0; // All P1.x reset
P2DIR = 0xFF; // All P2.x outputs
P2OUT = 0; // All P2.x reset
P3SEL = 0x30; // P3.4,5 = USCI_A0 TXD/RXD
P3DIR = 0xFF; // All P3.x outputs
P3OUT = 0; // All P3.x reset
UCA0CTL1 |= UCSSEL_1; // CLK = ACLK
UCA0BR0 = 0x03; // 32kHz/9600 = 3.41
UCA0BR1 = 0x00; //
UCA0MCTL = UCBRS1 + UCBRS0; // Modulation UCBRSx = 3
UCA0CTL1 &= ~UCSWRST; // **Initialize USCI state machine**
IE2 |= UCA0RXIE; // Enable USCI_A0 RX interrupt
__bis_SR_register(GIE);
//__bis_SR_register(LPM3_bits + GIE); // Enter LPM3 w/ int until Byte RXed
IE2 |= UCA0TXIE; // Enable USCI_A0 TX interrupt
UCA0TXBUF = string1[i++];
while(1);
}
#pragma vector=USCIAB0TX_VECTOR
__interrupt void USCI0TX_ISR(void)
{
UCA0TXBUF = string1[i++]; // TX next character//断点就设在这里怎么不进入呢我在main里已经从串口发东西了呀？
if (i == sizeof string1 - 1) // TX over?
IE2 &= ~UCA0TXIE; // Disable USCI_A0 TX interrupt
}
#pragma vector=USCIAB0RX_VECTOR
__interrupt void USCI0RX_ISR(void)
{
if (UCA0RXBUF == 'u') // 'u' received?
{
i = 0;
IE2 |= UCA0TXIE; // Enable USCI_A0 TX interrupt
UCA0TXBUF = string1[i++];
}
}
断点设在串口发送中断里软仿真不停在断点上 为什么 谢谢 |
楼主
标题置顶
标题高亮
