Multiple Port Interrupt on same Port

Mallappa T

Other Parts Discussed in Thread: MSP430G2553

Dear Sir,

Please help me to get multiple interrupt on Same Port, I have written code for Single port interrupt, And here i have inserted whatever i have written for port interrupt code.

#include <msp430g2553.h>

void main(void) {
   WDTCTL = WDTPW + WDTHOLD;            // Stop watchdog timer

   P1DIR |= BIT0;                    // Set P1.0 to output direction
   P1OUT &= ~BIT0;                // Set LED off
   P1SEL &= ~BIT3;                   // Select Port 1 P1.3 (push button)
   P1DIR &= ~BIT3;                   // Port 1 P1.3 (push button) as input, 0 is input
   P1REN |= BIT3;                       // Enable Port P1.3 (push button) pull-up resistor
   P1IE |= BIT3;                       // Port 1 Interrupt Enable P1.3 (push button)
   P1IFG &= ~BIT3;                   // Clear interrupt flag
   _BIS_SR(GIE);                          // Enable interrupts

}
                                   // Port 1 interrupt service routine
#pragma vector=PORT1_VECTOR
      __interrupt void Port_1(void) {

   P1IFG &= ~BIT3;                    // P1.3 Interrupt Flag cleared
   P1OUT ^= BIT0;                    // Toggle LED state

}

over 12 years ago

0 Jens-Michael Gross over 12 years ago

Guru 227245 points

What do you mean with 'multiple interrupts'? More than one port pin triggering an interrupt? Or more than one interrupt form the same pin?
Interrupts don't stack. Once an interrupt event has been detected, the IFG bit is set and remains set until cleared. If more events happen before it was cleared, you won't notice (well, some modules have an overflow detection, like the timer or the ADC)

For multiple interrupts on different pins of a port, each pin sets its own IFG bit. Inside the ISR check the IFG bits and handle the interrupts.

0 Leo Bosch over 12 years ago in reply to Jens-Michael Gross

Guru 23000 points

Or he has a bouncing or IS-edge problem.

You need to add: P1OUT |= BIT3; to make it a pull-up, also add P1IES |= BIT3; to react on a button-push (high-to-low).

0 Mallappa T over 12 years ago in reply to Jens-Michael Gross

Intellectual 295 points

Dear Sir,

I want multiple interrupt from same port but different pins, And also multiple interrupt on same pin.

This is the code i have written and it dint works only, red led toggle and when i press P1.4 that time also Red Led glows

#include <msp430g2553.h>

void main(void) {
   WDTCTL = WDTPW + WDTHOLD;            // Stop watchdog timer

   P1DIR |= BIT0;                    // Set P1.6 to output direction
   P1OUT &= ~BIT0;                // Set green LED off
   P2DIR |= BIT6;                    // Set P1.6 to output direction
   P2OUT &= ~BIT6;                // Set green LED off
   P1SEL &= ~BIT3;                   // Select Port 1 P1.3 (push button)
   P1DIR &= ~BIT3;                   // Port 1 P1.3 (push button) as input, 0 is input
   P1REN |= BIT3;                       // Enable Port P1.3 (push button) pull-up resistor
   P1IE |= BIT3;                       // Port 1 Interrupt Enable P1.3 (push button)
   P1IFG &= ~BIT3;                   // Clear interrupt flag
   P1SEL &= ~BIT4;                   // Select Port 1 P1.3 (push button)
   P1DIR &= ~BIT4;                   // Port 1 P1.3 (push button) as input, 0 is input
   P1REN |= BIT4;                       // Enable Port P1.3 (push button) pull-up resistor
   P1IE |= BIT4;                       // Port 1 Interrupt Enable P1.3 (push button)
   P1IFG &= ~BIT4;                   // Clear interrupt flag
   _BIS_SR(GIE);                          // Enable interrupts

}
                                   // Port 1 interrupt service routine
#pragma vector=PORT1_VECTOR
      __interrupt void Port_1(void) {

   P1IFG &= ~BIT3;                    // P1.3 Interrupt Flag cleared
   P1OUT ^= BIT0;                // Toggle LED state
   }
#pragma vector=PORT1_VECTOR
      __interrupt void Port_2(void) {
   P1IFG &= ~BIT4;                    // P1.3 Interrupt Flag cleared
   P1OUT ^= BIT6;                    // Toggle LED state
}

0 Jens-Michael Gross over 12 years ago in reply to Mallappa T

Guru 227245 points

Each port has only one vector and onyl one ISR can be written. To distinguish between different interrupt sources for one interrupt vector, in this case the different port pins, you'll have to check the IFG bits inside the ISR.

#pragma vector=PORT1_VECTOR
__interrupt void port1_ISR(void) {
if(P1IFG&BIT3)
P1OUT^=BIT0;
if(P1IFG&BIT4)
P1OUT ^=BIT6;
P1IFG=0;
}

On MSPs with P1 and P2 interrupt vector register, it might be better to use it:

#pragma vector=PORT1_VECTOR
__interrupt void port1_ISR(void) {
while(1)
    switch(_even_in_range(P1IV,16) { // check for highest pending interrupt and clear its IFG bit
      case 0: return; // no more pending interrupts
      case 8: // P1IFG.3
        P1OUT^=BIT0;
        break;
      case 16: //P1IFG.0
        P1OUT ^=BIT6;
        break;
    }
}
}

This will enter the ISR only once, even if multiple interrupts are pending, but it will only handle those interrupts where the IE bit is set and which had called the ISR anyway.

0 Mallappa T over 12 years ago in reply to Jens-Michael Gross

Intellectual 295 points

Dear Sir,

I didn't understand second part of the code, what does "_even_in_range(P1IV,16)" mean, P1IV is not in the header file.

Thanks for your information.

0 Jens-Michael Gross over 12 years ago in reply to Mallappa T

Guru 227245 points

That's why I wrote "on MSPs with P1 and P2 interrupt vector register" (I didn't remember what MSP you are using when writing the response).
The G2 family MSPs don't have IV registers for the ports, only for the timers and some other modules.

0 Mallappa T over 12 years ago in reply to Jens-Michael Gross

Intellectual 295 points

Thank You Sir....

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Multiple Port Interrupt on same Port