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.

Practice

1. Which function is used to set a pin as an output before writing HIGH or LOW to it?

easy

A. setPinMode(pin, HIGH)

B. digitalWrite(pin, OUTPUT)

C. pinMode(pin, OUTPUT)

D. writePin(pin, OUTPUT)

Solution

1. Step 1: Understand pin configuration

   Before writing HIGH or LOW, the pin must be set as an output to control voltage.

2. Step 2: Identify the correct function

   The function to set pin mode is pinMode(pin, OUTPUT), which configures the pin for output.

3. Final Answer:

   pinMode(pin, OUTPUT) -> Option C

4. Quick Check:

   Set pin as output = pinMode(pin, OUTPUT) [OK]

Hint: Set pin mode with pinMode before writing HIGH or LOW [OK]

Common Mistakes:

• Using digitalWrite to set pin mode
• Confusing HIGH/LOW with OUTPUT/INPUT
• Forgetting to set pin mode before writing

2. Which of the following is the correct syntax to turn pin 5 ON (HIGH) in embedded C?

easy

A. digitalWrite(5, 1);

B. digitalWrite(5, ON);

C. digitalWrite(5, OUTPUT);

D. digitalWrite(5, HIGH);

Solution

1. Step 1: Identify the function to write pin state

   The function digitalWrite(pin, value) sets the pin voltage HIGH or LOW.

2. Step 2: Use correct constants

   Use HIGH to turn the pin ON. The correct syntax is digitalWrite(5, HIGH);.

3. Final Answer:

   digitalWrite(5, HIGH); -> Option D

4. Quick Check:

   Turn pin ON = digitalWrite(pin, HIGH) [OK]

Hint: Use HIGH constant, not ON or numbers, for clarity [OK]

Common Mistakes:

• Using ON instead of HIGH
• Using OUTPUT instead of HIGH
• Using numeric 1 instead of HIGH constant

3. What will be the output state of pin 3 after running this code?

pinMode(3, OUTPUT);
digitalWrite(3, LOW);
digitalWrite(3, HIGH);

medium

A. Pin 3 will be LOW

B. Pin 3 will be HIGH

C. Pin 3 will be in input mode

D. Pin 3 will be undefined

Solution

1. Step 1: Set pin 3 as output

   The code sets pin 3 as output with pinMode(3, OUTPUT);.

2. Step 2: Write LOW then HIGH to pin 3

   First, pin 3 is set LOW, then immediately set HIGH, so final state is HIGH.

3. Final Answer:

   Pin 3 will be HIGH -> Option B

4. Quick Check:

   Last digitalWrite sets pin state = HIGH [OK]

Hint: Last digitalWrite sets the pin state [OK]

Common Mistakes:

• Assuming pin stays LOW after writing HIGH
• Forgetting pinMode sets output mode
• Thinking pin state is undefined without delay

4. Identify the error in this code snippet:

pinMode(7, OUTPUT);
digitalWrite(7, HIGH)
digitalWrite(7, LOW);

medium

A. Missing semicolon after digitalWrite(7, HIGH)

B. pinMode should be called after digitalWrite

C. HIGH and LOW are not valid constants

D. Pin number 7 cannot be used

Solution

1. Step 1: Check syntax line by line

   The second line digitalWrite(7, HIGH) is missing a semicolon at the end.

2. Step 2: Confirm other lines are correct

   pinMode(7, OUTPUT); and digitalWrite(7, LOW); are correct syntax.

3. Final Answer:

   Missing semicolon after digitalWrite(7, HIGH) -> Option A

4. Quick Check:

   Every statement must end with semicolon [OK]

Hint: Check each line ends with a semicolon [OK]

Common Mistakes:

• Forgetting semicolon after function calls
• Calling pinMode after digitalWrite
• Misusing HIGH/LOW constants

5. You want to toggle pin 10 ON and OFF every second in a loop. Which code snippet correctly sets pin 10 HIGH then LOW with a 1-second delay between states?

hard

A. pinMode(10, OUTPUT);
digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

B. digitalWrite(10, HIGH);
delay(1000);
digitalWrite(10, LOW);
delay(1000);

C. pinMode(10, INPUT);
digitalWrite(10, HIGH);
delay(1000);

D. pinMode(10, OUTPUT);
digitalWrite(10, LOW);
digitalWrite(10, HIGH);

Solution

1. Step 1: Set pin 10 as output

   Use pinMode(10, OUTPUT); to configure pin 10 for output.

2. Step 2: Write HIGH, wait, then LOW, wait

   Use digitalWrite(10, HIGH); then delay(1000); to wait 1 second, then digitalWrite(10, LOW); and delay again.

3. Final Answer:

   pinMode(10, OUTPUT); digitalWrite(10, HIGH); delay(1000); digitalWrite(10, LOW); delay(1000); -> Option A

4. Quick Check:

   Set output, write states with delay between [OK]

Hint: Always set pinMode before toggling with delays [OK]

Common Mistakes:

• Forgetting to set pinMode as OUTPUT
• Missing delay between HIGH and LOW
• Setting pinMode to INPUT instead of OUTPUT