Rising, falling, and change triggers in Arduino - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Detecting a Rising Edge Trigger

What will be the output on the serial monitor when the button connected to pin 2 is pressed (assuming it was previously not pressed)?

Arduino

const int buttonPin = 2;
int buttonState = 0;
int lastButtonState = 0;

void setup() {
  pinMode(buttonPin, INPUT);
  Serial.begin(9600);
}

void loop() {
  buttonState = digitalRead(buttonPin);
  if (buttonState == HIGH && lastButtonState == LOW) {
    Serial.println("Rising edge detected");
  }
  lastButtonState = buttonState;
  delay(50);
}

A Change detected

B Falling edge detected

C Rising edge detected

D No output

💡 Hint

Think about when the button state changes from LOW to HIGH.

The code prints "Rising edge detected" only when the button state changes from LOW to HIGH, which is a rising edge trigger.

❓ Predict Output

intermediate

Detecting a Falling Edge Trigger

What will be printed on the serial monitor when the button connected to pin 3 is pressed (assuming it was previously not pressed)?

Arduino

const int buttonPin = 3;
int buttonState = 0;
int lastButtonState = HIGH;

void setup() {
  pinMode(buttonPin, INPUT_PULLUP);
  Serial.begin(9600);
}

void loop() {
  buttonState = digitalRead(buttonPin);
  if (buttonState == LOW && lastButtonState == HIGH) {
    Serial.println("Falling edge detected");
  }
  lastButtonState = buttonState;
  delay(50);
}

A No output

B Rising edge detected

C Change detected

D Falling edge detected

💡 Hint

Focus on when the button state changes from HIGH to LOW.

The code prints "Falling edge detected" only when the button state changes from HIGH to LOW, which is a falling edge trigger.

❓ Predict Output

advanced

Detecting Any Change Trigger

What will be printed on the serial monitor when the button connected to pin 4 changes state (pressed or released)?

Arduino

const int buttonPin = 4;
int buttonState = 0;
int lastButtonState = 0;

void setup() {
  pinMode(buttonPin, INPUT);
  Serial.begin(9600);
}

void loop() {
  buttonState = digitalRead(buttonPin);
  if (buttonState != lastButtonState) {
    Serial.println("Change detected");
  }
  lastButtonState = buttonState;
  delay(50);
}

A Change detected

B No output

C Falling edge detected

D Rising edge detected

💡 Hint

Think about any difference between current and last button state.

The code prints "Change detected" whenever the button state changes, either from LOW to HIGH or HIGH to LOW.

🧠 Conceptual

advanced

Understanding Interrupts with Rising and Falling Edges

Which statement correctly describes how to use interrupts to detect a falling edge on pin 2 in Arduino?

A attachInterrupt(digitalPinToInterrupt(2), ISR, RISING);

B attachInterrupt(digitalPinToInterrupt(2), ISR, FALLING);

C attachInterrupt(2, ISR, CHANGE);

D attachInterrupt(digitalPinToInterrupt(2), ISR, HIGH);

💡 Hint

Look for the correct interrupt mode for falling edge detection.

To detect a falling edge, the interrupt mode must be FALLING. The function digitalPinToInterrupt converts the pin number to the interrupt number.

❓ Predict Output

expert

Output of Combined Edge Detection with Interrupts

Given the code below, what will be the output on the serial monitor if the button connected to pin 3 is pressed and then released?

Arduino

volatile int count = 0;

void ISR() {
  count++;
}

void setup() {
  pinMode(3, INPUT_PULLUP);
  Serial.begin(9600);
  attachInterrupt(digitalPinToInterrupt(3), ISR, CHANGE);
}

void loop() {
  static int lastCount = 0;
  if (count != lastCount) {
    Serial.print("Interrupt count: ");
    Serial.println(count);
    lastCount = count;
  }
}

A Interrupt count: 2

B Interrupt count: 1

C Interrupt count: 0

D No output

💡 Hint

Each press and release triggers one interrupt with CHANGE mode.

The CHANGE mode triggers an interrupt on both rising and falling edges. Pressing and releasing the button causes two interrupts, so count increments twice.