attachInterrupt() function in Arduino - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Arduino sketch using attachInterrupt()?

Consider the following Arduino code snippet. What will be printed on the Serial Monitor after pressing the button connected to pin 2?

Arduino

volatile int count = 0;

void setup() {
  Serial.begin(9600);
  pinMode(2, INPUT_PULLUP);
  attachInterrupt(digitalPinToInterrupt(2), incrementCount, FALLING);
}

void loop() {
  Serial.println(count);
  delay(1000);
}

void incrementCount() {
  count++;
}

A The Serial Monitor prints random numbers due to interrupt noise.

B The Serial Monitor prints increasing numbers every second starting from 0, incrementing each time the button is pressed.

C The Serial Monitor prints 0 repeatedly and never changes.

D The Arduino resets continuously causing no output.

💡 Hint

Think about what happens when the button is pressed and how the interrupt updates the variable.

The interrupt triggers on the FALLING edge of pin 2, calling incrementCount() which increases count. The loop prints count every second, so the number increases each time the button is pressed.

🧠 Conceptual

intermediate

Which statement about attachInterrupt() is true?

Choose the correct statement about the attachInterrupt() function in Arduino.

A <code>attachInterrupt()</code> allows you to specify a function to run when a pin changes state without polling in <code>loop()</code>.

B <code>attachInterrupt()</code> can only be used with pins 0 and 1 on all Arduino boards.

C The interrupt service routine (ISR) can safely use <code>delay()</code> to wait inside the function.

D You must call <code>detachInterrupt()</code> before using <code>attachInterrupt()</code> again on the same pin.

💡 Hint

Think about what interrupts do compared to normal code flow.

attachInterrupt() lets you run a function automatically when a pin changes state, without checking it repeatedly in loop(). Pins vary by board, ISRs should be short and not use delay(), and detachInterrupt() is optional.

🔧 Debug

advanced

Why does this attachInterrupt() code cause a compile error?

Identify the cause of the compile error in this Arduino code snippet:

Arduino

void setup() {
  attachInterrupt(2, blinkLED(), RISING);
}

void blinkLED() {
  digitalWrite(13, HIGH);
  delay(100);
  digitalWrite(13, LOW);
}

A The function blinkLED() must return an int, not void.

B The pin number 2 is invalid for attachInterrupt() on Arduino Uno.

C The ISR blinkLED() uses delay(), which is not allowed inside interrupts.

D The interrupt function blinkLED() is called immediately instead of passing its name as a pointer.

💡 Hint

Look carefully at how the function is passed to attachInterrupt().

The code calls blinkLED() immediately by using parentheses. Instead, the function name without parentheses should be passed to attachInterrupt() to register it as the ISR.

📝 Syntax

advanced

Which option correctly uses attachInterrupt() to trigger on a falling edge?

Choose the correct syntax to attach an interrupt on pin 3 that triggers on a falling edge and calls the function handleInterrupt.

A attachInterrupt(3, handleInterrupt, FALLING);

B attachInterrupt(3, handleInterrupt(), FALLING);

C attachInterrupt(digitalPinToInterrupt(3), handleInterrupt, FALLING);

D attachInterrupt(digitalPinToInterrupt(3), handleInterrupt(), FALLING);

💡 Hint

Remember how to convert a pin number to an interrupt number and how to pass the function.

Option C correctly uses digitalPinToInterrupt(3) to get the interrupt number and passes the function name handleInterrupt without parentheses.

🚀 Application

expert

How many times will the variable 'count' increment in this scenario?

Given this Arduino code, if a noisy button connected to pin 2 is pressed once causing multiple rapid FALLING edges, how many times will count increment after the press?

Arduino

volatile int count = 0;

void setup() {
  pinMode(2, INPUT_PULLUP);
  attachInterrupt(digitalPinToInterrupt(2), incrementCount, FALLING);
}

void loop() {}

void incrementCount() {
  count++;
}

A Multiple times, because each FALLING edge triggers the interrupt, including noise.

B Count will increment randomly due to debounce hardware.

C Zero times, because interrupts are disabled by default.

D Exactly 1 time, because the interrupt triggers only once per press.

💡 Hint

Think about how mechanical button noise affects interrupts triggered on edges.

Mechanical buttons often cause multiple rapid transitions (bouncing) when pressed. Each FALLING edge triggers the interrupt, so count increments multiple times unless debouncing is implemented.