Setting a specific bit in a register in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the value of the register after setting bit 3?

Given a register initially set to 0x00, what will be its value after setting bit 3 using the code below?

Embedded C

unsigned char reg = 0x00;
reg |= (1 << 3);
printf("0x%02X", reg);

A 0x08

B 0x03

C 0x10

D 0x80

💡 Hint

Remember that bits are zero-indexed from the right, starting at 0.

Setting bit 3 means turning on the 4th bit from the right (0-based). This corresponds to the value 0x08 in hexadecimal.

❓ Predict Output

intermediate

What does this code print after setting bit 0 and bit 7?

Consider the following code snippet. What is printed?

Embedded C

unsigned char reg = 0x00;
reg |= (1 << 0);
reg |= (1 << 7);
printf("0x%02X", reg);

A 0x80

B 0x01

C 0x81

D 0xFF

💡 Hint

Bit 0 is the least significant bit, bit 7 is the most significant bit in an 8-bit register.

Setting bit 0 adds 0x01, setting bit 7 adds 0x80. Together they form 0x81.

🔧 Debug

advanced

Identify the error in setting bit 5

What error does this code produce when trying to set bit 5 of the register?

Embedded C

unsigned char reg = 0x00;
reg = reg | 1 << 5;
printf("0x%02X", reg);

A No error, prints 0x20

B Syntax error due to missing parentheses

C Runtime error due to operator precedence

D Prints 0x01 instead of 0x20

💡 Hint

Check operator precedence of bitwise OR and shift operators.

The bit shift operator has higher precedence than bitwise OR, so 1 << 5 is evaluated first, then ORed with reg. This sets bit 5 correctly.

📝 Syntax

advanced

Which option correctly sets bit 4 without affecting other bits?

Choose the code snippet that sets bit 4 of reg without changing other bits.

Embedded C

unsigned char reg = 0x12; // initial value

A reg |= 1 << 5;

B reg = reg & (1 << 4);

C reg = (1 << 4);

D reg = reg | (1 << 4);

💡 Hint

Use bitwise OR to set a bit without clearing others.

Option D uses OR to set bit 4, preserving other bits. Option D clears bits except bit 4. Option D overwrites reg. Option D sets bit 5, not bit 4.

🚀 Application

expert

How many bits are set after this code runs?

Given the code below, how many bits are set to 1 in reg after execution?

Embedded C

unsigned char reg = 0x00;
for (int i = 0; i < 8; i += 2) {
    reg |= (1 << i);
}
printf("%d", __builtin_popcount(reg));

A 5

B 4

C 8

D 3

💡 Hint

Bits set are at positions 0, 2, 4, 6.

The loop sets bits 0, 2, 4, and 6. That is 4 bits set in total.