unable to use PIC18 low isr

Started by embe...@austinlakes.com July 3, 2008
I have been attempting to use the PIC18 low isr but can't seem to get anything but the high isr to fire. I have reduced my code to the following and it works fine as long as I use the high isr. If I set a break point on the low one, it never gets accessed. The original code was designed to use the low isr only and I have been able to make it work by pointing the high vector to the low handler. Changing IPEN makes no difference. Not really what I wanted!

Any suggestions?

#pragma config OSC = HS
#pragma config PWRT = OFF
#pragma config BOREN = OFF
#pragma config WDT = OFF
#pragma config MCLRE = ON
#pragma config PBADEN = OFF
#pragma config LVP = OFF
void timer0_isr(void); // Interrupt

void init(void); // Initial

#pragma code high_vector=0x08

void high_vector(void)
_asm GOTO timer0_isr _endasm

#pragma code /* return to the default code section */

#pragma code low_vector=0x18

void low_vector(void)
_asm GOTO timer0_isr _endasm

#pragma code /* return to the default code section */
//#pragma interruptlow timer0_isr

void timer0_isr (void)
unsigned int i;



INTCONbits.TMR0IF = 0;
INTCONbits.GIE = 1; //jb added when seen that GIE gets reset

void main (void)

// timer0_isr(); //jb added as test to display temp prior to ints

OpenTimer0 (TIMER_INT_ON & T0_SOURCE_INT & T0_16BIT);
RCONbits.IPEN=1; //jb added due to c18 app note on no high isr
INTCON2bits.TMR0IP = 1;
INTCON2bits.RBIP = 0;
INTCONbits.TMR0IF = 0; //jb added to clear the tmr0 flag

INTCONbits.GIE = 1;