2 little questions

Sascha W.

Hello,

First i have a problem with a simple program. (I have the same in a bigger code, so first i tried it in a smaller one..)

First code with if:

Fullscreen 8032.aiste.txt Download

#include <msp430G2153.h>

void main(void){

	WDTCTL = WDTPW | WDTHOLD;

	BCSCTL1 = CALBC1_16MHZ;
	DCOCTL = CALDCO_16MHZ;
	
	P1DIR |= BIT0;
	P1SEL = 0x00;

	while(1){
		if(P1IN & BIT1)
			P1OUT |=BIT0;
		else
			P1OUT &= ~BIT0;
	}
}

The picture from osc.:

Second code with an interrupt.

Fullscreen 3771.int.txt Download

#include <msp430G2153.h>

#pragma vector=PORT1_VECTOR
__interrupt void Port_1(void)
{
	P1OUT ^= BIT0;
	P1IFG &= ~BIT2;
	P1IES ^= BIT2;
}

void main(void){

	WDTCTL = WDTPW | WDTHOLD;

	BCSCTL1 = CALBC1_16MHZ;
	DCOCTL = CALDCO_16MHZ;

	P1DIR |= BIT0;
	P1SEL = 0x00;
	P1IFG &= ~BIT2;
	P1IE |= BIT2;
	_enable_interrupts ();

	while(1){
	}
}

The picture from the osc.:

My Question: Why is there a little interrupt after the low-high edge?

My second question is, when i will compile my code in ccs5 there occurs an error:

"10010 errrors encountered during linking; project.out not built

"#10099-D program will not fit into"

It is only when i have a "if" with more than 2 conditions like: if(a==0 && b==1 && c<=5){}

I am very thankfull for any answers ;)

Best regards

over 13 years ago

0 Bruce McKenney over 13 years ago

Guru 100905 points

I suspect you're tripping over the conditions in section 8.2.7.2 in the User Guide. Summary of that table seems to be that if you request "A->B edge" in the IES and the pin is already at "B", you might get an immediate (stray-ish) interrupt.

The indicated fix would be to set (change) the IES Before clearing the IFG.

----

Out of context, I'm not sure what those CCS messages are telling you.

0 Sascha W. over 13 years ago in reply to Bruce McKenney

Prodigy 250 points

Hello,

thank you for your answer, but the problem is unfortunately not solves.

When i change the settings of ifg like this:

Fullscreen 2465.int2.txt Download

#include <msp430G2153.h>

#pragma vector=PORT1_VECTOR
__interrupt void Port_1(void)
{
	P1OUT ^= BIT0;
	P1IES ^= BIT2;
	P1IFG &= ~BIT2;
}

void main(void){

	WDTCTL = WDTPW | WDTHOLD;

	BCSCTL1 = CALBC1_16MHZ;
	DCOCTL = CALDCO_16MHZ;

	P1DIR |= BIT0;
	P1SEL = 0x00;
	P1IFG &= ~BIT2;
	P1IE |= BIT2;
	_enable_interrupts ();

	while(1){
	}
}

Now first i change the ies, then i delete the ifg.

But then i have the following signal:

0 Vaibhav Ostwal5 over 13 years ago in reply to Sascha W.

TI__Expert 6160 points

hi,

by writing this statement

P1IES ^= BIT2;

you are changing the edge on which the interrupt occurs every time the ISR is called.

in other words, by a single button press you are triggering 2 interrupts '

1. one on high- to low transition

2. the other on low-to high transition

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2 little questions