Configuring pin as input or output in Embedded C - Practice Exercises

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this pin configuration code?

Consider a microcontroller where setting a bit in the DDR (Data Direction Register) configures the pin as output (1) or input (0). What will be the value of DDRB after running this code?

Embedded C

DDRB = 0x00;
DDRB |= (1 << 3);  // Set pin 3 as output
DDRB &= ~(1 << 1); // Set pin 1 as input

A 0x0A

B 0x08

C 0x0C

D 0x02

💡 Hint

Remember that (1 << 3) sets bit 3 and (1 << 1) sets bit 1. |= sets bits, &=~ clears bits.

Initially DDRB is 0x00 (all pins input). Setting bit 3 (pin 3) to 1 makes it output. Clearing bit 1 keeps it input. So only bit 3 is set, which is 0x08 in hex.

🧠 Conceptual

intermediate

Which statement correctly configures pin 5 as input?

Given a DDR register controlling pin directions, which code snippet correctly sets pin 5 as input without changing other pins?

A DDR &= ~(1 << 5);

B DDR = (1 << 5);

C DDR |= (1 << 5);

D DDR ^= (1 << 5);

💡 Hint

Input pins have DDR bit cleared (0). Use bitwise AND with NOT to clear a bit.

To set pin 5 as input, clear bit 5 in DDR. DDR &= ~(1 << 5) clears bit 5 and keeps others unchanged.

🔧 Debug

advanced

Why does this code fail to configure pin 2 as output?

This code is intended to set pin 2 as output by setting bit 2 in DDR. Why does it not work as expected? DDRB = 0; DDRB & (1 << 2);

A DDRB is reset to 0 after setting the bit.

B The bit shift is incorrect; it should be (1 << 3).

C The code uses & instead of |=, so it does not set the bit.

D Pin 2 cannot be configured as output on this microcontroller.

💡 Hint

Check the operator used to set bits.

Using & does not set bits; it performs bitwise AND. To set a bit, use |= operator.

📝 Syntax

advanced

Which option contains a syntax error in configuring pin 0 as output?

Identify the code snippet that will cause a syntax error when trying to set pin 0 as output.

A DDRB |= (1 << 0);

B DDRB = DDRB | (1 << 0);

C DDRB |= 1 << 0;

D DDRB |= (1 << 0

💡 Hint

Look for missing parentheses or incomplete expressions.

Option D is missing a closing parenthesis, causing a syntax error.

🚀 Application

expert

How many pins are configured as output after this code runs?

Given DDRC initially 0x00, what is the number of pins configured as output after running this code? DDRC |= (1 << 0); DDRC |= (1 << 2); DDRC &= ~(1 << 0); DDRC |= (1 << 7);

A 2

B 3

C 1

D 4

💡 Hint

Count bits set to 1 in DDRC after all operations.

Pin 0 is first set then cleared (input). Pins 2 and 7 are set as output. So total 2 pins output.

Practice

1. Which register is used in embedded C to configure a microcontroller pin as input or output?

easy

A. PORT (Data Register)

B. TIMER Register

C. PIN (Input Register)

D. DDR (Data Direction Register)

Solution

1. Step 1: Understand pin configuration registers

   The DDR register controls whether a pin is input or output by setting bits to 0 or 1.

2. Step 2: Identify the correct register for direction

   PORT controls output values, PIN reads input values, TIMER is unrelated to pin direction.

3. Final Answer:

   DDR (Data Direction Register) -> Option D

4. Quick Check:

   DDR sets pin direction [OK]

Hint: Remember DDR sets direction: 1=output, 0=input [OK]

Common Mistakes:

• Confusing PORT with DDR
• Using PIN register to set direction
• Thinking TIMER controls pin mode

2. Which of the following lines correctly sets pin 3 of PORTB as output in embedded C?

easy

A. DDRB |= (1 << 3);

B. DDRB &= ~(1 << 3);

C. PORTB |= (1 << 3);

D. PINB |= (1 << 3);

Solution

1. Step 1: Understand setting pin as output

   To set a pin as output, set its bit in DDR register to 1 using bitwise OR.

2. Step 2: Analyze each option

   DDRB |= (1 << 3); uses DDRB |= (1 << 3) which sets bit 3 to 1 correctly. DDRB &= ~(1 << 3); clears bit 3 (input). Options C and D affect PORTB and PINB, not direction.

3. Final Answer:

   DDRB |= (1 << 3); -> Option A

4. Quick Check:

   Use |= with DDR to set output [OK]

Hint: Use |= with DDR to set output bit [OK]

Common Mistakes:

• Using &= ~ instead of |= to set output
• Modifying PORT or PIN instead of DDR
• Shifting wrong bit position

3. What is the effect of the following code snippet?

DDRC |= (1 << 2);  // Line 1
DDRC &= ~(1 << 2); // Line 2

medium

A. Pin 2 of PORTC is set as output then input

B. Pin 2 of PORTC remains output

C. Pin 2 of PORTC remains input

D. Syntax error in code

Solution

1. Step 1: Analyze Line 1

   DDRC |= (1 << 2) sets bit 2 to 1, configuring pin 2 as output.

2. Step 2: Analyze Line 2

   DDRC &= ~(1 << 2) clears bit 2 to 0, configuring pin 2 as input.

3. Final Answer:

   Pin 2 of PORTC is set as output then input -> Option A

4. Quick Check:

   Set then clear bit changes pin direction [OK]

Hint: |= sets output, &=~ clears to input [OK]

Common Mistakes:

• Thinking both lines set output
• Ignoring second line effect
• Confusing PORTC with DDRC

4. Identify the error in this code that tries to set pin 5 of PORTD as input:

DDRD |= (0 << 5);

medium

A. DDRD cannot be modified directly

B. Bit shift is incorrect, should be (1 << 5)

C. Should use &= ~ instead of |= to set input

D. No error, code is correct

Solution

1. Step 1: Understand setting pin as input

   To set a pin as input, clear its bit in DDR register using &= ~ (bitwise AND with NOT).

2. Step 2: Analyze the code

   DDRD |= (0 << 5) does nothing because (0 << 5) is 0; it does not clear bit 5. Correct way is DDRD &= ~(1 << 5).

3. Final Answer:

   Should use &= ~ instead of |= to set input -> Option C

4. Quick Check:

   Use &= ~ to clear bit for input [OK]

Hint: Use &= ~ to clear bit for input, not |= [OK]

Common Mistakes:

• Using |= with 0 does not clear bit
• Confusing bitwise OR and AND
• Not using bitwise NOT (~) when clearing

5. You want to configure pins 0, 1, and 2 of PORTA as outputs and pins 3 and 4 as inputs without changing other pins. Which code snippet correctly does this?

hard

A. DDRA |= (1 << 0) | (1 << 1) | (1 << 2) | (1 << 3) | (1 << 4);

B. DDRA |= (1 << 0) | (1 << 1) | (1 << 2); DDRA &= ~((1 << 3) | (1 << 4));

C. DDRA |= (1 << 0) & (1 << 1) & (1 << 2); DDRA &= ~((1 << 3) & (1 << 4));

D. DDRA = (1 << 0) | (1 << 1) | (1 << 2); DDRA = ~((1 << 3) | (1 << 4));

Solution

1. Step 1: Set pins 0,1,2 as outputs

   Use bitwise OR to set bits 0,1,2 without affecting others: DDRA |= (1<<0)|(1<<1)|(1<<2);

2. Step 2: Set pins 3,4 as inputs

   Clear bits 3 and 4 using bitwise AND with NOT: DDRA &= ~((1<<3)|(1<<4));

3. Final Answer:

   DDRA |= (1 << 0) | (1 << 1) | (1 << 2); DDRA &= ~((1 << 3) | (1 << 4)); -> Option B

4. Quick Check:

   Use |= to set output bits, &=~ to clear input bits [OK]

Hint: Set outputs with |=, inputs with &= ~ combined masks [OK]

Common Mistakes:

• Overwriting DDRA instead of modifying bits
• Using & instead of | for combining bits
• Setting input pins as output by mistake