Multiple Arduino Interrupts on same pin

Question

In the following code, why does ISR2 never run?

const int in = 19;
const int LED = 9;
int flag;
void setup() {
    pinMode(in, INPUT_PULLUP);
    pinMode(LED, OUTPUT);
    attachInterrupt(digitalPinToInterrupt(in), ISR2, RISING);
    attachInterrupt(digitalPinToInterrupt(in), ISR1, FALLING);

}
void  loop() {
    if (flag) {delay (100); flag = false;}// debounce
}
void ISR1(){
    digitalWrite(LED, LOW);
    // Turn Off the motor, since, Limit Switch was engaged
    flag = true;
}
void ISR2(){ // Limit Switch was disengaged. 
    digitalWrite(LED, HIGH);
    delay(100); // Debounce, so I do not receive spurious FALLING edges from this.
}

Does the Arduino not allow you to attach two interrupts on the same pin even if the interrupts are programmed for different events?

In my setup, pin 19 gets a signal from a limit switch used in a motion control setup. When the limit switch is engaged, the in pin gets LOW signal. Thus, I first see a FALLING edge followed by RISING edges and FALLING edges due to mechanical bounce. I handle the debouncing correctly in this case.

However, imagine the Limit Switch was sitting in engaged state for a while, and then I reverse the motor causing the Limit Switch to disengage, this will send a RISING edge followed by FALLING and RISING edges. I need to ignore these edges since nothing is in danger. The ISR2 was written with purpose of capturing the first RISING edge when the Limit switch disengages, and then debouncing it so that the following FALLING edges are ignored. But now that ISR2 never gets called, how can I deal with this situation?

P.S. My microcontroller is ATMEGA 2650, It's a Arduino Mega board.

Without looking at the AVR datasheet, I'd suspect you can't set up two interrupts on the same pin. It seems to be universal on all the platforms I've worked on that you get one interrupt source per pin, and you can trigger edge, rising, level, falling, etc. — Russ Schultz
Using busy-waiting in an ISR is evil! Check the source code of delay() I already commented this on your last question. However, I do not know if Arduino covers that in software, but what you do looks very strange and is ... uncommon. AVR (like all other MCUs I know) cannot use two interrupts on the same pin. Please learn some more about how interrupts and hardware work. And how to debounce a switch properly. Your approach is definitively wrong. — too honest for this site

kkrambo kkrambo · Accepted Answer · 2015-10-28T15:22:36

ISR2 never runs because there can be only one interrupt service routine for any interrupt source and you have replaced ISR2 with ISR1 as the sercice routine for this interrupt. If you reordered your code and attached ISR1 before ISR2 then you might see ISR2 run but not ISR1.

A typical microcontroller has an interrupt vector table that associates an interrupt service routine with each interrupt source. There can be only one service routine for each interrupt source. If you assign an new service routine then you're replacing the old service routine. There are not separate service routines for the rising and falling edges. Rising versus falling edge is a configuration setting for the interrupt source to determine when that interrupt should fire. The interrupt source cannot be configured to fire on both edges simultaneously.

However you may be able to reconfigure the interrupt for the other edge after you have received the interrupt for the first edge and debounced the transition. This way your code will ping-pong back and forth configuring for one ISR and then the other.

Multiple Arduino Interrupts on same pin

2 Answers