GPIO port-wide operations in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this GPIO port write operation?

Given a 8-bit GPIO port initially set to 0x00, what will be the port value after executing the following code?

Embedded C

uint8_t GPIO_PORT = 0x00;

// Set pins 0, 2, and 4 high
GPIO_PORT |= (1 << 0) | (1 << 2) | (1 << 4);

printf("0x%02X", GPIO_PORT);

A 0x15

B 0x1A

C 0x14

D 0x16

💡 Hint

Remember that (1 << n) sets bit n to 1. Combine bits with OR.

Bits 0, 2, and 4 correspond to values 1, 4, and 16 respectively. ORing them gives 1 + 4 + 16 = 21 decimal, which is 0x15 in hex.

❓ Predict Output

intermediate

What is the value of GPIO_PORT after clearing bits?

Starting with GPIO_PORT = 0xFF, what is the value after clearing bits 1, 3, and 5 using this code?

Embedded C

uint8_t GPIO_PORT = 0xFF;

// Clear bits 1, 3, and 5
GPIO_PORT &= ~((1 << 1) | (1 << 3) | (1 << 5));

printf("0x%02X", GPIO_PORT);

A 0xAB

B 0xD5

C 0xBF

D 0x7F

💡 Hint

Use bitwise AND with the negation of the bits to clear.

Bits 1, 3, and 5 correspond to 0x02, 0x08, and 0x20. Clearing these bits from 0xFF (all ones) results in 0xFF & ~(0x2A) = 0xFF & 0xD5 = 0xAB.

🔧 Debug

advanced

Why does this GPIO port toggle code not work as expected?

This code is intended to toggle bits 0 and 7 of GPIO_PORT. What is the problem?

Embedded C

uint8_t GPIO_PORT = 0xAA; // 0b10101010

// Toggle bits 0 and 7
GPIO_PORT ^= 1 << 0 & 1 << 7;

printf("0x%02X", GPIO_PORT);

A The code toggles all bits instead of just bits 0 and 7.

B The bitwise AND has higher precedence than XOR, so only bit 0 is toggled incorrectly.

C The code causes a syntax error due to missing parentheses.

D The code toggles bits 0 and 7 correctly, output is 0x2A.

💡 Hint

Check operator precedence between ^ and &.

The expression '1 << 0 & 1 << 7' is evaluated as '(1 << 0) & (1 << 7)' due to operator precedence, but in C, bitwise AND (&) has higher precedence than bitwise XOR (^), so the expression is evaluated as '1 << (0 & 1) << 7', which is incorrect. The bitwise AND has higher precedence than XOR, so only bit 0 is toggled incorrectly.

❓ Predict Output

advanced

What is the output after setting and clearing multiple GPIO port bits?

Given GPIO_PORT initially 0x55, what is the value after this sequence?

Embedded C

uint8_t GPIO_PORT = 0x55; // 0b01010101

// Set bits 1 and 3
GPIO_PORT |= (1 << 1) | (1 << 3);

// Clear bits 0 and 4
GPIO_PORT &= ~((1 << 0) | (1 << 4));

printf("0x%02X", GPIO_PORT);

A 0x6B

B 0x4A

C 0x5A

D 0x7B

💡 Hint

Apply set bits first, then clear bits.

Starting with 0x55 (01010101), setting bits 1 and 3 sets bits 1 and 3 to 1, resulting in 0x5F (01011111). Clearing bits 0 and 4 clears bits 0 and 4, resulting in 0x5A (01011010).

🧠 Conceptual

expert

How many bits are set in GPIO_PORT after this operation?

If GPIO_PORT is 0xF0 and you execute GPIO_PORT ^= 0xFF;, how many bits are set (1) in GPIO_PORT afterwards?

Embedded C

uint8_t GPIO_PORT = 0xF0;
GPIO_PORT ^= 0xFF;
// Count bits set in GPIO_PORT

A 6

B 8

C 0

D 4

💡 Hint

XOR with 0xFF flips all bits.

0xF0 is 11110000 in binary. XOR with 0xFF flips all bits, resulting in 00001111, which has 4 bits set.

Practice

1. What does writing a value to PORTB in embedded C typically do in GPIO port-wide operations?

easy

A. Configures the direction of pins in port B

B. Sets the output levels of all pins in port B at once

C. Reads the input levels of all pins in port B

D. Enables pull-up resistors on port B pins

Solution

1. Step 1: Understand PORT register role

   The PORT register controls the output voltage levels on the pins of a GPIO port.

2. Step 2: Effect of writing to PORTB

   Writing a value to PORTB sets the output state (high or low) of all pins in port B simultaneously.

3. Final Answer:

   Sets the output levels of all pins in port B at once -> Option B

4. Quick Check:

   PORTB write = set output levels [OK]

Hint: PORTx controls output levels of all pins together [OK]

Common Mistakes:

• Confusing PORT with DDR (direction register)
• Thinking PORT reads input levels
• Assuming PORT configures pull-ups

2. Which of the following is the correct syntax to set all pins of port D as outputs in embedded C?

easy

A. PINB = 0xFF;

B. PORTD = 0xFF;

C. DDRD = 0xFF;

D. DDRB = 0x00;

Solution

1. Step 1: Identify DDR register purpose

   DDRx registers set pin directions: 1 for output, 0 for input.

2. Step 2: Set all pins of port D as output

   Writing 0xFF (all bits 1) to DDRD sets all pins as outputs.

3. Final Answer:

   DDRD = 0xFF; -> Option C

4. Quick Check:

   DDRD = 0xFF sets all outputs [OK]

Hint: Use DDRx = 0xFF to set all pins output [OK]

Common Mistakes:

• Using PORTD instead of DDRD to set direction
• Writing to PINB which is read-only
• Setting DDRB instead of DDRD

3. Given the code snippet:

DDRC = 0x00;
PORTC = 0xFF;
uint8_t val = PINC;

What is the value of val assuming all pins on port C are externally connected to ground?

medium

A. 0x00

B. Undefined behavior

C. 0xFF with pull-ups disabled

D. 0xFF

Solution

1. Step 1: Analyze DDRC and PORTC settings

   DDRC = 0x00 sets all pins as inputs. PORTC = 0xFF enables pull-up resistors on all pins.

2. Step 2: Effect of external ground on input pins

   Even with pull-ups enabled, external ground forces pins low, so PINC reads 0x00.

3. Final Answer:

   0x00 -> Option A

4. Quick Check:

   Input pins grounded read 0x00 [OK]

Hint: Input pins grounded read zero despite pull-ups [OK]

Common Mistakes:

• Assuming pull-ups force input high regardless
• Confusing PORTC write with output level
• Ignoring DDRC input configuration

4. Identify the error in the following code snippet for setting port A pins as outputs and turning them all high:

DDRA = 0xFF;
PINA = 0xFF;

medium

A. PORTA should be used instead of PINA to set outputs

B. DDRA should be set to 0x00

C. PINA is not a valid register name

D. No error, code is correct

Solution

1. Step 1: Check register purpose

   PINA reads input levels or toggles outputs, but does not directly set output levels.

2. Step 2: Correct register to set output levels

   To set output pins high, write to PORTA, not PINx.

3. Final Answer:

   PORTA should be used instead of PINA to set outputs -> Option A

4. Quick Check:

   Use PORTA to write outputs, not PINx [OK]

Hint: Use PORTx to write outputs, PINx is for reading inputs [OK]

Common Mistakes:

• Using PINx registers instead of PORTx for outputs
• Writing outputs to PIN register
• Confusing DDR and PORT registers

5. You want to configure port C pins 0-3 as outputs and pins 4-7 as inputs with pull-ups enabled. Which code correctly achieves this?

hard

A. DDRC = 0x00; PORTC = 0xFF;

B. DDRC = 0xF0; PORTC = 0x0F;

C. DDRC = 0xFF; PORTC = 0x00;

D. DDRC = 0x0F; PORTC = 0xF0;

Solution

1. Step 1: Set pins 0-3 as outputs

   Bits 0-3 set to 1 in DDRC means DDRC = 0x0F (00001111 binary).

2. Step 2: Enable pull-ups on pins 4-7 inputs

   Pins 4-7 inputs require DDRC bits 4-7 = 0, and PORTC bits 4-7 = 1 to enable pull-ups, so PORTC = 0xF0.

3. Final Answer:

   DDRC = 0x0F; PORTC = 0xF0; -> Option D

4. Quick Check:

   Outputs low nibble, pull-ups high nibble [OK]

Hint: DDRx bits 1=output, PORTx bits 1=pull-up on inputs [OK]

Common Mistakes:

• Mixing up which bits control direction vs pull-ups
• Setting pull-ups on output pins
• Using wrong hex values for bit masks