Ivan,
Try replacing the 1 with a 0. This will place the ISR at 0x0003, where
the PIC expects it.
void isr_high(void) interrupt 0 {
...
Joel
On Jan 28, 2008 12:06 PM, Ivan Petrushev <[EMAIL PROTECTED]> wrote:
> Hello,
> I'm new to SDCC and microcontrollers - my first two very simple
> programs burned into PIC16F88 and working! :)
> Now I'm trying something little harder than just lighting LEDs - I
> want to try the interrupt system.
> PIC16F88 has one external interrupt - on pin RB0.
> I'm trying the following very simple program, but something is wrong:
> #define __16f88
> #include"pic/pic16f88.h"
>
> typedef unsigned int word;
> word at 0x2007 __CONFIG = 0x3f70;
>
> void isr_high(void) interrupt 1 {
> if (RB0 == 1) RA1 = 0;
> }
> void main(void) {
> TRISA = 0;
> while(1) {
> RA1 = 1;
> }
> }
>
> I expect when I run this to have pin RA1 '1', untill RB0 is pulled
> '1'. Then RA1 should go '0' untill RB0 is released.
> I'm running that at gpsim and there RA1 is always '1'. When I run the
> program step by step I see it never enters the interrupt code section.
> So, how can I define an interrupt? I've searched a lot through the
> documentation of SDCC and the only piece of value I've found is the
> 'void isr_high interrupt 1' function I'm using.
>
> Is there a more complicated/simplified way to code an interrupt?
>
> Thanks,
> Ivan.
>
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