Assume that the condition is satisfied for an interrupt to occur and interrupt is enabled along with I bit. What will happen if ISR is not written? Is the behavior is undefined or ISR will just return without executing?
I one of my program, I observed strange behavior. In the outline code given below, I had forgot to implement ISR(PCINT2_vect). Because of that code was getting executed until while(!flag) and it was waiting for the flag to set; but then rest of the code was getting skipped. I was trying to use PCINT2 interrupt, but forgot to implement. I was expecting endless wait because flag was not getting set inside ISR, but while(1) was getting entered again and again
while(1)
{
//Implement first Part
while(!flag);
// Implement second part
}
inline void commonCode(void)
{
// some code here common to all ISRs
}
ISR(PCINT0_vect)
{
flag=1;
commonCode();
}
ISR(PCINT1_vect)
{
flag=1;
commonCode();
}
ISR(PCINT2_vect)
{
flag=1;
commonCode();
}Secondly, if I have some common code to be executed by ISR, is it OK to write an inline function and call it from ISR like the one shown above?
