Experiment with both CC registers for Timer A on the MSP430G2231

timerA.c

#include <msp430g2211.h>


#define LED_0 BIT0
#define LED_1 BIT6
#define LED_OUT P1OUT
#define LED_DIR P1DIR



unsigned int timerCount = 0;
unsigned int timerCount1 = 0;
void main(void)
{
	WDTCTL = WDTPW + WDTHOLD; // Stop watchdog timer
	LED_DIR |= (LED_0 | LED_1); // Set P1.0 and P1.6 to output direction
	LED_OUT &= ~(LED_0 | LED_1)       ; // Set the LEDs off
	TACCTL0 = CCIE;
	TACCTL1 = CCIE;
	TACTL = TASSEL_2 | MC_2; // Set the timer A to SMCLCK, Continuous
	// Clear the timer and enable timer interrupt
	
	__enable_interrupt();
	
	__bis_SR_register(LPM0 + GIE); // LPM0 with interrupts enabled

}


// Timer A0 interrupt service routine
#pragma vector=TIMERA0_VECTOR
__interrupt void Timer_A0 (void)
{
	timerCount = (timerCount + 1) % 8;
	if(timerCount ==0)
		P1OUT ^= (LED_0);
}

// Timer A1 interrupt service routine
#pragma vector=TIMERA1_VECTOR
__interrupt void Timer_A1 (void)
{
	TAIV &= ~(0x02); // CLEAR INTERRUPT
	timerCount1 = (timerCount1 + 1) % 9; //4096;
	if(timerCount1 ==0)
		P1OUT ^= (LED_1);
}

Thanks for your reply. You explained it clearly. I now understand the purpose of the modulus operator and how it is applied in the code.