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.