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 7 times, with each button press occurring at regular 200ms 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 7 times, with each button press occurring at regular 200ms 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 boolean interruptFired is set true by any interrupt and is cleared only when loop() checks it, so regardless of how many button presses happen during a single 1 s delay, the code will only increment counter once per loop iteration.

1) Relevant knowledge points and concepts
- attachInterrupt sets an ISR that runs asynchronously and sets interruptFired = true on each falling edge.
- interruptFired is a boolean flag (not a counter), so multiple interrupts between checks only leave it true (not accumulated).
- loop() contains delay(1000); the code checks interruptFired once after each 1 s delay and, if true, increments counter by exactly 1 and clears the flag.

2) Step-by-step timing reasoning
- Time origin: loop() starts its first iteration at t = 0 and immediately calls delay(1000) which lasts until t = 1000 ms.
- Button presses occur every 200 ms starting at t = 5 ms. So the seven press times are:
  Press 1: t = 5 ms
  Press 2: t = 205 ms
  Press 3: t = 405 ms
  Press 4: t = 605 ms
  Press 5: t = 805 ms
  Press 6: t = 1005 ms
  Press 7: t = 1205 ms

3) Intermediate steps / calculations
- During the first delay interval (0–1000 ms) presses 1–5 occur. Each sets interruptFired = true, but the flag remains true; no multiple increments can occur until checked.
- At t ≈ 1000 ms the delay ends, loop() checks interruptFired: it is true (due to presses 1–5), so counter = 0 + 1 = 1, then interruptFired is set false.
- The second delay runs from 1000–2000 ms. Presses 6 and 7 (1005 ms and 1205 ms) occur in this interval, setting interruptFired true (again only once effectively).
- At t ≈ 2000 ms the loop checks interruptFired again: true, so counter = 1 + 1 = 2.

4) Why the correct answer is correct
Because the code uses a boolean flag that only records whether at least one interrupt occurred since the last check, each 1-second loop iteration can only add one to counter regardless of how many presses happen in that second. The first 5 presses fall in the first second and count as one increment; the last 2 presses fall in the second second and count as another increment.

5) Why other plausible answers are incorrect
- "7": assumes every press increments counter immediately — incorrect because the flag is only read once per second.
- "5" or other values: may count presses per grouping incorrectly; the correct grouping is 5 presses in the first interval → 1 increment, 2 in the next → 1 increment.

6) Conclusion
The final value of counter is 2, which matches the provided answer.

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