Interrupt volatile variable doesn't work as expected

Question

Please examine the following code.

volatile uint8_t myVar;

void setup() {
  attachInterrupt(digitalPinToInterrupt(2), isr, FALLING);
  Serial.begin(115200);
}

void loop() {
  myVar = 0;
  while (myVar != 5) {
  }
  Serial.println("ISR called");
}

void isr() {
  myVar = 5;
}

From this code, I expect that every time the pin 2 is falling, I get one single print statement ISR called.

However, when dropping the voltage on pin 2 ONCE, the serial monitor shows ISR called TWICE.

Why is this? This is a simplified, stripped-down example of the problem I am facing with my current project.

Edit: After reading the comments, here is a bit more info. I am using a 64-button shield that is triggering the interrupt. The process is time critical since I am measuring the time it takes for the button to be pressed (I am testing people's reflexes). I will add a print statement in the interrupt to see how many times it gets called. However, I believe the interrupt is only getting called once per buttonPressed.

Thank you for your help!

Answer 1

Turns out, the 64 button shield calls the interrupt TWICE. Once when the button is pressed and once when it is released.

Thank you for everyone's help!