MSP430FG4619--RTC

The RTC is based on 32768Hz clock, which won’t divide well into 1/1000s (but divides well with a simple binary divider). You won’t get ms resolution form the RTC.

Hello professor,

now, I take your advice to use timer module to test, but I found the code cannot jump out of the interrupt, could you help me to have a look about my configuration?  Is there something wrong in it? thanks so much for your time

void Init_TIMERA0(void)
{
  TA0CCR0 = 1000;  //count limit (16 bit) 12000
  TA0CCTL0 = CCIE;  //enable counter interrupts, bit 4=1, TACCR0 interrupt enabled
  TA0CTL = TASSEL_2 + MC_1;  //SMCLK
 
}


#pragma vector = TIMER0_A1_VECTOR
__interrupt void Timer0_A1 (void)
{
  Time_miliseconds();
  //timer++;
}

#pragma vector = TIMER0_A0_VECTOR
__interrupt void Timer0_A0 (void)
{
  //Time_miliseconds();
  //timer++;
  asm("nop");
}

int main( void )
{
    WDTCTL = WDTPW + WDTHOLD;
   InitFreq_XT2();
   Init_TIMERA0();
   __bis_SR_register(GIE);
while(1)
{
  DrawMenu();   //the pointer stops here and it cannot go down anymore
..........
.........
}
}

The interrupt for TA0CCR0 jumps to TIMER0_A0_VECTOR.
All other interrupts of TimerA0 jump to TIMER0_A1_VECTOR.

However, you'll need to manually clear the IFG bits for those other interrupts. Only CCIE in TA0CCR0 clears by itself (as it is the only trigger for this interrupt).
If you doN't clear the IFG bits (whether directly, or by reading the TA0IV register), the ISR will be called over and over again and main execution will effectively stop.

But the code you posted should call TIMER0_A0_VECTOR ISR and execute the nop in it.

What does InitFreq() do? Or TimeMilliseconds()? (BTW: it is not the best idea to call functions form within an ISR. It adds lots of overhead, and perhaps the function does something that stalls the system (if it isn't aware that it is called form within an ISR)