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Message |
feederic
Joined: 08 Apr 2008
Posts: 7
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| PortB Interrupts and Interrupt Flags |
 Posted: Sat Apr 19, 2008 7:05 pm |
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I'm trying to use the portb interrupts using RB0-RB4 on the 16f887.
I'm not sure if there are flags for each pin that I can check. Basically i want my code to do something like this. Each button will output a LED on interrupt. I've tried the code below, but after an interrupt is generated all the flags are true.
I'm not sure if each pin on port b already has its own interrupt flag that I can check or if I have to write my own.
| Code: |
#INT_RB
void button_interrupt_service(void)
{
if(int_rb0)
{
while(1)
output_d(0xff);
}
if(int_rb1)
{
while(1)
output_d(1);
}
if(int_rb2)
{
while(1)
output_d(2);
}
if(int_rb3)
{
while(1)
output_d(4);
}
if(int_rb4)
{
while(1)
output_d(8);
}
}
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Please offer any advice or helpful links. |
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foodwatch
Joined: 18 Apr 2006
Posts: 66
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| Posted: Sat Apr 19, 2008 8:55 pm |
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| If memory serves me correctly, the interrupt occurs when ANY of the four inputs change. You must then quickly read the pins with your interrupt routine to see which one really did change. |
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Ttelmah
Guest
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| Posted: Sun Apr 20, 2008 2:45 am |
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The existing reply is spot on.
There is an internal 'latch' in the chip, reflecting the state of the inputs, _when they were last read_. The interrupt occurs, when the input pins differ from this state. Inside the interrupt, you _must_ read the port (to reset this latch). It is up to you to determine which pin changed. Also remember that the interrupts will occur on both the rising and falling edges of the signals. A typical 'RB changed' handler, would be:
| Code: |
static int8 old_pins;
#define first_bit(x) (x&0x1)
#define second_bit(x) (x&0x2)
#define third_bit(x) (x&0x4)
#define fourth_bit(x) (x&0x8)
#INT_RB
void b_changed(void) {
int new_pins;
int changes;
new_pins=input_b(); //read the port to reset the latch
//Now simply test the bits, to see which pin(s)
//have changed.
changes=new_pins^old_pins; //This now reflects every bit that has
//changed.
changes=changes&new_pins; //This now reflects every bit that has
//gone _high_ (use 'old_pins', instead of 'new_pins',if you want to find
//the ones that have gone low).
old_pins=new_pins; //update your stored value for the future
if (first_bit(changes)) {
//here RB0 has gone high
}
if (second_bit(changes)) {
//Here RB1 has gone high
}
if (third_bit(changes)) {
//Here RB2 has gone high
}
if (fourth_bit(changes)) {
//Here RB3 has gone high
}
}
//Inside the 'main' at the start of the code, just before enabling the
//interrupt have:
old_pins=input_b();
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There are individual flags in this chip to _enable_ which pins will generate the interrupt, but nothing to say which pin has triggered the interrupt. If you look at Figure 14-7 in the data sheet, you will see that there is only one interrupt 'flag', fed by all these sources (top left corner of the diagram).
Best Wishes |
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feederic
Joined: 08 Apr 2008
Posts: 7
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| Posted: Mon Apr 21, 2008 12:56 am |
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Thanks for the advice! The buttons are now working perfectly!
I'm glad I stumbled on to this forum, I've learned more this past week than in the past two months  ! |
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