Digital Switch strange results

I am not always printing the message “switch open interrupt!” when I open a closed switch.
I use D6 and set it high and connect to D7 via a digital closed switch. With this setup the D7 blue led is on, but getting strange results when opening switch multiple times. Is this a correct setup to monitor a closed switch and generate an interrupt when it goes open?

int switch = D7;
int sethigh3v = D6;

void setup() {
    pinMode(sethigh3v, OUTPUT);
    digitalWrite(sethigh3v, HIGH);  
    pinMode(switch, INPUT_PULLDOWN);    
    attachInterrupt(switch, isrEvent, FALLING);

void loop() {
	//check digital switch

void isrEvent() { 
    if (digitalRead(button)==HIGH && systemHasNoticed == false) {     
      systemHasNoticed = true;   
void checkEvents() {   
 if(digitalRead(button)==HIGH && systemHasNoticed == true) {    
     Serial.println("switch open interrupt!");
     systemHasNoticed = false;    

When triggering on a falling edge you’d usually use INPUT_PULLUP and close the switch to GND.

Also we can’t see your declaration of systemHasNoticed - this boolean should be marked volatile.


Yes declaration of systemHasNoticed is marked volatile, but I changed the
attachInterrupt(switch, isrEvent, FALLING); to attachInterrupt(switch, isrEvent, RISING);
and using switch as normally open rather than normally closed.

This seems to have cleared up the issue.

thanks for your help.


Thanks for the assist!

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