The ADC12 interrupt service routine.

I found out that i still can get the ADC result without the switch case function like the coding shown below:

#pragma vector=ADC12_VECTOR
__interrupt void ADC12ISR (void)
{
  static unsigned int index = 0;

   A0results[index] = ADC12MEM0;           // Move A0 results, IFG is cleared
    A1results[index] = ADC12MEM1;           // Move A1 results, IFG is cleared
    A2results[index] = ADC12MEM2;           // Move A2 results, IFG is cleared
    A3results[index] = ADC12MEM3;           // Move A3 results, IFG is cleared
    index++;                                // Increment results index, modulo; Set Breakpoint1 here
   
    if (index == 8)
    {
      (index = 0);
    }

}

Then what's the point of using switch case in the sample coding?

The ADC12 interrupt is a shared interrupt. So you need to read the index to determine who generated the interrupt. If you only have one interrupt enabled, you probably don't need to use the switch/case function. But I would still use an if statement to check for the specific index you are looking for so that other interrupts or a zero index don't mess you up. 

Timothy Barr said:

The ADC12 interrupt is a shared interrupt. So you need to read the index to determine who generated the interrupt. If you only have one interrupt enabled, you probably don't need to use the switch/case function. But I would still use an if statement to check for the specific index you are looking for so that other interrupts or a zero index don't mess you up. 

Thanks for the comments.

Is the ISR that I mentioned above a forever loop? Which means it doesn't go back to the main() routine anymore? Or it occasionally will goes back to the main() routine? Below is the main() routine for your review:

void main(void)
{
  WDTCTL = WDTPW+WDTHOLD;                   // Stop watchdog timer
  P6SEL = 0x0F;                             // Enable A/D channel inputs
  ADC12CTL0 = ADC12ON+ADC12MSC+ADC12SHT0_8; // Turn on ADC12, extend sampling time
                                            // to avoid overflow of results
  ADC12CTL1 = ADC12SHP+ADC12CONSEQ_3;       // Use sampling timer, repeated sequence
  ADC12MCTL0 = ADC12INCH_0;                 // ref+=AVcc, channel = A0
  ADC12MCTL1 = ADC12INCH_1;                 // ref+=AVcc, channel = A1
  ADC12MCTL2 = ADC12INCH_2;                 // ref+=AVcc, channel = A2
  ADC12MCTL3 = ADC12INCH_3+ADC12EOS;        // ref+=AVcc, channel = A3, end seq.
  ADC12IE = 0x08;                           // Enable ADC12IFG.3
  ADC12CTL0 |= ADC12ENC;                    // Enable conversions
  ADC12CTL0 |= ADC12SC;                     // Start convn - software trigger
 
  __bis_SR_register(LPM0_bits + GIE);       // Enter LPM0, Enable interrupts
  __no_operation();                         // For debugger
 
}

So, will it exit the ISR and goes back to the main() routine? Or it stays in the ISR as a forever loop?

That depends on how fast the interrupts are triggered, if the interrupts occur faster than you execute the ISR, then the MCU will stay in the ISR (it will exit it and reenter it immediately as there is a pending interrupt). If you can handle the ISR fast enough it will return to the code where the interrupt occurred. In your case your main method is not doing anything but the initialization and go to sleep then. So even if it enters main() again, it won't do anything you will notice.

Steven Gan Teck Yik said:

I found out that i still can get the ADC result without the switch case function like the coding shown below:

Yes and no. The switch/case structure ensure sthat exactly the values is processed that has triggered the ISR.

In your second version, you assume that all ADC12MEMx contain a new value and the ISR was triggered by the last. Which is a totally valid, but completely different approach, which has advantages and disadvantages depending on the exact application.