ID:q4vC. Study the following diagram and code:

boolean light_on = false;
boolean interruptFired = false;
int counter = 0;

void setup() {
Serial.begin(9600);
attachInterrupt(0, interruptServiceRoutine, FALLING);
pinMode(10, OUTPUT);
pinMode(11, OUTPUT);
pinMode(12, OUTPUT);
pinMode(13, OUTPUT);
pinMode(2, INPUT);
}

void loop() {
delay(1000);
if (interruptFired) {
interruptFired = false;
counter = counter + 1;
Serial.print("Current Count ");
Serial.print(counter);
Serial.println(";");
}
digitalWrite(10, bitRead(counter, 0));
digitalWrite(11, bitRead(counter, 1));
digitalWrite(12, bitRead(counter, 2));
digitalWrite(13, bitRead(counter, 3));
}

void interruptServiceRoutine() {
interruptFired = true;
}

This circuit is intended to count each button press, displaying the current count on the terminal and in unsigned binary on the LEDs.

Suppose the button is pressed 5 times, with each button press occurring at regular 500ms intervals. The first button press occurs 5 ms after the loop function starts executing for the first time (5 ms into the delay(1000) line). You can neglect the time it takes to execute statements other than 'delay()' in the function 'void loop()'.

What is the final value of the variable counter? Write your answer in decimal.

Note that the code and question are different from your tutorial activities.

Question

ID:q4vC. Study the following diagram and code:

boolean light_on = false;
boolean interruptFired = false;
int counter = 0;

void setup() {
  Serial.begin(9600);
  attachInterrupt(0, interruptServiceRoutine, FALLING);
  pinMode(10, OUTPUT);
  pinMode(11, OUTPUT);
  pinMode(12, OUTPUT);
  pinMode(13, OUTPUT);
  pinMode(2, INPUT);
}

void loop() {
  delay(1000);
  if (interruptFired) {
    interruptFired = false;
    counter = counter + 1;        
    Serial.print("Current Count ");
    Serial.print(counter);
    Serial.println(";");
  }
  digitalWrite(10, bitRead(counter, 0));
  digitalWrite(11, bitRead(counter, 1));
  digitalWrite(12, bitRead(counter, 2));
  digitalWrite(13, bitRead(counter, 3));
}

void interruptServiceRoutine() {
  interruptFired = true;
}

This circuit is intended to count each button press, displaying the current count on the terminal and in unsigned binary on the LEDs.

Suppose the button is pressed 5 times, with each button press occurring at regular 500ms intervals. The first button press occurs 5 ms after the loop function starts executing for the first time (5 ms into the delay(1000) line). You can neglect the time it takes to execute statements other than 'delay()' in the function 'void loop()'.

What is the final value of the variable counter? Write your answer in decimal.

Note that the code and question are different from your tutorial activities.

BlackTom AI · Accepted Answer

The ISR only sets interruptFired = true and the loop samples that flag once per iteration (after a 1000 ms delay), so multiple button presses that occur during the same delay interval produce only one increment.

1) Relevant knowledge points
- attachInterrupt triggers the ISR on each falling edge; the ISR sets a single boolean flag (interruptFired = true).
- loop() begins with delay(1000). The code only increments counter when it sees interruptFired true, then clears the flag.
- Any number of interrupts between checks will leave the flag true (no queuing or counting of multiple presses).

2) Step-by-step timing and reasoning
- Time t = 0 ms: loop() starts the first delay(1000) (covers 0–1000 ms).
- Press times (given): press1 at 5 ms, press2 at 505 ms, press3 at 1005 ms, press4 at 1505 ms, press5 at 2005 ms.
- During first delay (0–1000 ms): press1 at 5 ms sets interruptFired = true; press2 at 505 ms sets interruptFired = true again (no change). At t = 1000 ms, delay ends and loop checks the flag once → sees true, sets interruptFired = false and executes counter = counter + 1 → counter becomes 1.
- Next loop iteration delay (1000–2000 ms): press3 (1005 ms) and press4 (1505 ms) both occur during this delay and set the flag (again coalesced). At t = 2000 ms the loop checks and increments once → counter becomes 2.
- Next loop iteration delay (2000–3000 ms): press5 at 2005 ms sets the flag. At t = 3000 ms the loop checks and increments once → counter becomes 3.

3) Intermediate calculations shown as events → increments: after first check = 1, after second check = 2, after third check = 3.

4) Why this is correct
Because the ISR does not count presses, it only marks that at least one press occurred since the last check. The code increments counter once per loop iteration when the flag is true, so groups of presses within a single 1000 ms delay produce a single increment.

5) Why other counts would be wrong
- 5 would assume every press directly increments counter immediately (not true; increments happen only when loop samples the flag).
- 2 or 4 would mis-group the presses into different delay windows; the timing above shows three separate sampling moments capturing the five presses as three groups.

6) Conclusion
Final value of counter is 3, which matches the provided answer.

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