Writing HIGH and LOW to output pins in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output state of the pin after this code?

Consider this embedded C code snippet that sets a pin high and then low. What is the final state of the pin?

Embedded C

#define PIN 5
#define HIGH 1
#define LOW 0

int pin_state = LOW;

void digitalWrite(int pin, int state) {
    if (pin == PIN) {
        pin_state = state;
    }
}

int main() {
    digitalWrite(PIN, HIGH);
    digitalWrite(PIN, LOW);
    return pin_state;
}

A 1

B 0

C -1

D Compilation error

💡 Hint

Think about the last command that sets the pin state.

The pin is first set HIGH (1), then immediately set LOW (0). The final state is LOW (0).

🧠 Conceptual

intermediate

Which statement correctly sets a pin HIGH in embedded C?

You want to set a digital output pin to HIGH. Which of the following code lines correctly does this?

A digitalWrite(pin, HIGH);

B digitalWrite(pin);

C digitalWrite(pin, 2);

D digitalWrite(HIGH, pin);

💡 Hint

The function takes the pin number first, then the state.

The function digitalWrite expects the pin number first, then the state (HIGH or LOW). Only option A follows this.

🔧 Debug

advanced

Why does this code fail to set the pin LOW?

This code is intended to set the pin LOW but it does not work as expected. What is the error?

Embedded C

#define PIN 3
#define HIGH 1
#define LOW 0

int pin_state = HIGH;

void digitalWrite(int pin, int state) {
    if (pin = PIN) {
        pin_state = state;
    }
}

int main() {
    digitalWrite(PIN, LOW);
    return pin_state;
}

A LOW is not defined properly.

B The pin_state variable is not declared correctly.

C The if condition uses assignment '=' instead of comparison '==', causing always true.

D digitalWrite function is missing a return statement.

💡 Hint

Check the if condition inside digitalWrite carefully.

The if condition uses '=' which assigns PIN to pin, always true, so pin_state is set. But this can cause unexpected behavior if pin is changed.

❓ Predict Output

advanced

What is the output of this pin toggle code?

This code toggles a pin state starting from LOW. What is the final pin state after running main?

Embedded C

#define PIN 2
#define HIGH 1
#define LOW 0

int pin_state = LOW;

void digitalWrite(int pin, int state) {
    if (pin == PIN) {
        pin_state = state;
    }
}

void togglePin(int pin) {
    if (pin_state == LOW) {
        digitalWrite(pin, HIGH);
    } else {
        digitalWrite(pin, LOW);
    }
}

int main() {
    togglePin(PIN);
    togglePin(PIN);
    togglePin(PIN);
    return pin_state;
}

A 2

B 0

C Compilation error

D 1

💡 Hint

Count how many times the pin toggles starting from LOW.

Starting LOW, togglePin changes state: LOW->HIGH (1), HIGH->LOW (2), LOW->HIGH (3). Final state is HIGH (1).

🧠 Conceptual

expert

Which option causes a compilation error when writing HIGH or LOW to a pin?

Which of these code snippets will cause a compilation error when trying to write HIGH or LOW to a pin?

A digitalWrite(PIN, "HIGH");

B digitalWrite(PIN, HIGH);

C digitalWrite(PIN, 1);

D digitalWrite(PIN, true);

💡 Hint

Consider the data types expected by digitalWrite.

digitalWrite expects an integer for the state. Passing a string like "HIGH" causes a compilation error.