How to count using Interrupts

Question

I have an issue where I want to count how many times I hit a switch using an interrupt. I tried placing the count variable inside the main but it claims that count does not exist. I put count inside the pragma but I know that will reset the count value to 0. How should I approach this problem?

I'm using the STM8S003K3

volatile int count = 0;

#pragma vector = 8
__interrupt void EXTI_PORTD_IRQHandler(void){   
   count++;
   PD_ODR_ODR3 = !PD_ODR_ODR3;
}

void main() {
   __disable_interrupt();
   PD_ODR = 0;             //  All pins are turned off.
   PD_DDR = 0xff;          //  All pins except PD4 are outputs.
   PD_CR1 = 0xff;          //  Push-Pull outputs.
   PD_CR2 = 0xff;          //  Output speeds up to 10 MHz.
   //
   //  Now configure the input pin.
   //
   PD_DDR_DDR4 = 0;        //  PD4 is input.
   PD_CR1_C14 = 0;         //  PD4 is floating input.
   //
   //  Set up the interrupt.
   //
   EXTI_CR1_PDIS = 2;      //  Interrupt on falling edge only.
   EXTI_CR2_TLIS = 0;      //  Falling edge only.
   __enable_interrupt();

   while (1) {
       __wait_for_interrupt();
   }
}

Answer 1

Make it a volatile global variable; declare and initialize it outside all functions. Then you increment within your ISR and print the modified value in main(). Something like:

static volatile int count = 0;

isr(){
  count++;
...
}

void main(){
  debug_print(count);
...
}

The volatile keyword is to prevent your compiler from optimizing away the variable, since its not seen to change in the normal program flow.

EDIT: Declaring the variable static, like Lundin suggested, restricts its scope to this particular file, so that the variable is not visible outside this source file. You should use this only if the ISR and main() are defined within the same file.