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sotodaniel
Joined: 23 Jun 2015
Posts: 3
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| int_rb interrupts do not work |
 Posted: Tue May 03, 2016 7:27 pm |
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Greetings Ing. I wrote this program and my pc-1 compiles fine without any error or warnings. When programmed in the PIC16F887, and installed in the breadboard, it runs the program to perfection. However on another computer-2, I write this same program and it compiles fine without any errors or warnings. Then I program the PIC16F887 and put it in the breadboard and oh surprise, it does nothing ! or does not accept the interrupts. What is causing this ? Does the same program not work for different versions of CCS C ? Thank you for your help.
| Code: |
#include <UNO_10.h>
#use delay(clock=4000000)
#fuses XT, NOWDT, NOPROTECT, PUT, NOLVP
#use fast_io(b)
void parpadeo (void)
{
#BYTE TRISA = 0X85
TRISA = 0X00;
set_tris_a(0X00);
int i=0;
for(i=0; i<12; i++)
{
output_high(PIN_A1);
delay_ms(500);
output_low(PIN_A1);
delay_ms(500);
}
output_high(PIN_A1);
}
// standard i/o is the default.
// it is easier to handle interrupt on change with the pins
// fixed regarding their i/o, hence selecting fast_io.
int1 flag_int_RB4 = FALSE;
// flags to say these pin have_dropped_
//this sets each flag if the pin changes_low_, when it was high
#int_RB
void RB_isr(void)
{
static int1 old_RB4=1;
//handle RB4
if(input(PIN_B4)) //PIN is high
old_RB4=1; //record this
else
{
if(old_RB4==1) //pin is now low, and was high
{ //RB4 has changed to low
old_RB4=0;
flag_int_RB4=TRUE;
}
}
}
void main()
{
int temp;
set_tris_b(0XF0); //since you want to interrupt on B4, and B7
//they_must_be (u) inputs (/u);
port_B_pullups(0xFF);
setup_comparator(NC_NC_NC_NC);// This device COMP currently not supported by the PICWizard
temp=input_b(); //you must read the port to reset the change latch
clear_interrupt(INT_RB);
enable_interrupts(INT_RB);
enable_interrupts(GLOBAL);
while(TRUE)
{
//TODO: User Code
if(flag_int_RB4)
{
flag_int_RB4=FALSE;
//RB4 has now drppped -do what you want
//
///////////////
parpadeo();
//
///////////////
}
///////////// regreso de interrupciones
}///llave de cierre del while
}////llave de cierre del main |
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Ttelmah
Joined: 11 Mar 2010
Posts: 19550
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| Posted: Wed May 04, 2016 12:58 am |
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First tidy things. Make your indenting logical. Then there are silly things like setting TRIS twice. Unnecessary anyway, since you are not using fast_io on port a. Then use the C standard of putting all variable declarations at the start of the code sections. This is not _required_ by CCS on current versions, but is required on older versions, can give problems, even on current compilers, but especially on older compilers.
| Code: |
#include <UNO_10.h>
#use delay(clock=4000000)
#fuses XT, NOWDT, NOPROTECT, PUT, NOLVP
#use fast_io(b)
int1 flag_int_RB4 = FALSE;
void parpadeo (void)
{
int i=0;
for(i=0; i<12; i++)
{
output_high(PIN_A1);
delay_ms(500);
output_low(PIN_A1);
delay_ms(500);
}
output_high(PIN_A1);
}
// standard i/o is the default.
// it is easier to handle interrupt on change with the pins
// fixed regarding their i/o, hence selecting fast_io.
#int_RB
void RB_isr(void)
{
static int1 old_RB4=1;
//handle RB4
if(input(PIN_B4)) //PIN is high
old_RB4=1; //record this
else
{
if(old_RB4==1) //pin is now low, and was high
{ //RB4 has changed to low
old_RB4=0;
flag_int_RB4=TRUE;
}
}
}
void main(void)
{
int temp;
set_tris_b(0XF0); //since you want to interrupt on B4, and B7
//they_must_be (u) inputs (/u);
port_B_pullups(0xFF);
setup_comparator(NC_NC_NC_NC);
temp=input_b(); //you must read the port to reset the change latch
clear_interrupt(INT_RB);
enable_interrupts(INT_RB);
enable_interrupts(GLOBAL);
while(TRUE)
{
if(flag_int_RB4)
{
flag_int_RB4=FALSE;
//RB4 has now drppped -do what you want
parpadeo();
}
}///llave de cierre del while
}////llave de cierre del main
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If this doesn't work, then check your hardware. Are you sure RB4 actually is changing?. Are you sure the code was for the same chip?. On some modern PIC's you must use individual bit input to clear the latch, while a few older chips require byte reads to clear the latch. If the 887 is one where this applies, then the interrupt code will never exit.
Generally, you can run CCS code pretty universally, unless you are talking major changes (like between a V3 program and a V4 program, where some tweaking will apply). However even with this level of change, something as simple as this ought to work with just about any compiler version.
Are you sure the chip actually is starting?. Could be something as silly as an incorrect power connection, or MCLR not pulled up.... |
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