Interrupt works only once - PIC16F877a

Question

In my code, I have two interruptions, one is coming from the overflow of the TMR0, and the other one is when a button is pressed.

this is the code in MikroC :

int compt = 0;
int seconds = 10 ;
int enable = 0;

void interrupt(){

     if (INTCON.INTF) {
        PORTD = 9;
        enable = 1;
        seconds = 10;
        INTCON.INTF = 0;
     }

     if (INTCON.TMR0IF) {
        compt++;
        INTCON.TMR0IF  = 0;
        TMR0 = 0x06;
     }
}


void main() {

     TRISB = 0x01;
     PORTB = 0;

     PORTD = 0;
     TRISD = 0x00;


     INTCON = 0xB0;
     OPTION_REG = 0x44;
     TMR0 = 0x06;

     while(1){

        if (compt == 625){
           if (enable) seconds--;
           compt = 0;
        }

        if (seconds > 0 && enable == 1) {
           PORTD = seconds;
           PORTB.RB1 = 1;
        }  else {
            enable = 0;
            PORTB.RB1 = 0;
            PORTD = 0;
        }

     }

}

what I am trying to achieve with my code is as the following :

When I press one of the push buttons, the countdown starts and the LED illuminates until the countdown ends, and if the user pressed the button while the countdown still didn't hit 0, it starts over, until the countdown hits 0 again, then the LED should turn off.

What I'm facing here, is that the interruption from IRB0 works only once, the second time I press the button, nothing happens.

I am not sure if the TMR0F has something to do with that or not, tried many things, but couldn't make it to work.

I Hope that you could see something i didn't notice, and help me.

Answer 1

Many problems.

• You need to de-bounce the button. This is mentioned in any embedded beginner learning material.
• You should never connect a button directly to an interrupt-triggered pin without taking precautions. See https://stackoverflow.com/questions/23557921/how-to-detect-two-or-more-button-press-gpio-at-the-same-time-by-a-microprocess/23559522#23559522
• Variables shared between an ISR and the caller code must be volatile and protected against race conditions. See Using volatile in embedded C development.
• You need to be careful with how much current you sink/source from the pins. This particular part lists sink/source 20mA as absolute maximum ratings, the stress value. Depending on the forward voltage of the LEDs, you are close to that current with 220R. And what about the 7 seg? Is there a buffer somewhere? You need series resistors or a driver buffer.

Answer 2

There are some low level problems with your code that have to be handled first. These problems have the potential to break your code even if the logic would otherwise seem correct.

• Variables that are shared between the interrupt service routine (ISR) and the main program code should be declared volatile. In your case that is all three of them:

  volatile int compt = 0;
  volatile int seconds = 10;
  volatile int enable = 0;
  

Volatile tells the compiler that a variable can change in between different accesses even if the code does not appear to modify it. This prevents certain reads and writes being optimized away because the optimizer does not understand there's an ISR working with those same variables asynchronously.

• The C standard requires a int to be at least 16-bit. Your PIC, however, is a 8-bit device. This means manipulating a int will be a non-atomic operation. Non-atomic operations require multiple CPU instructions to complete. A interrupt could be triggered in the middle of such a operation which would cause either the ISR to see a (corrupt) intermediate value of the variable or the ISR modifying the variable while the main code was in the middle of doing an operation on it, possibly corrupting it.

A possible solution to this is to use a mutex or turn off interrupts during any relevant non-atomic operation.