Register bit manipulation patterns in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this register bit set operation?

Given a register REG initially set to 0x0F, what is the value of REG after setting bit 5?

Embedded C

unsigned char REG = 0x0F;
REG |= (1 << 5);
printf("0x%02X", REG);

A 0x2F

B 0x1F

C 0x3F

D 0x0F

💡 Hint

Remember that 1 << 5 sets bit 5 (counting from 0).

Starting with 0x0F (00001111), setting bit 5 (bit mask 00100000) results in 00101111 which is 0x2F.

❓ Predict Output

intermediate

What is the output after clearing a bit in a register?

Given REG = 0xFF, what is the value of REG after clearing bit 3?

Embedded C

unsigned char REG = 0xFF;
REG &= ~(1 << 3);
printf("0x%02X", REG);

A 0xEF

B 0xFB

C 0x7F

D 0xF7

💡 Hint

Clearing a bit means setting it to 0 using AND with the inverse mask.

Bit 3 mask is 00001000 (0x08). Inverting gives 11110111 (0xF7). AND with 0xFF clears bit 3, resulting in 0xF7.

❓ Predict Output

advanced

What is the output after toggling bits in a register?

Given REG = 0x55 (binary 01010101), what is the value of REG after toggling bits 0, 2, and 4?

Embedded C

unsigned char REG = 0x55;
REG ^= (1 << 0) | (1 << 2) | (1 << 4);
printf("0x%02X", REG);

A 0x40

B 0x15

C 0x55

D 0x75

💡 Hint

Toggling flips bits: 1 becomes 0, 0 becomes 1.

Bits 0,2,4 mask is 00010101 (0x15). XOR with 0x55 (01010101) flips those bits resulting in 01000000 (0x40).

❓ Predict Output

advanced

What is the output after checking a bit in a register?

Given REG = 0xA0, what is the output of the following code?

Embedded C

unsigned char REG = 0xA0;
if (REG & (1 << 7))
    printf("Set");
else
    printf("Clear");

A Clear

B Set

C Compile error

D No output

💡 Hint

Bit 7 corresponds to the highest bit in an 8-bit register.

0xA0 in binary is 10100000. Bit 7 is set (1), so the condition is true and "Set" is printed.

🧠 Conceptual

expert

How many bits are set in this register after the operation?

Given REG = 0x3C, and the operation REG = (REG << 1) | (REG >> 7); (a rotate left by 1 bit), how many bits are set (1) in REG after this operation?

Embedded C

unsigned char REG = 0x3C;
REG = (REG << 1) | (REG >> 7);
// Count bits set in REG

A 3

B 5

C 4

D 6

💡 Hint

Rotate left shifts bits left and wraps the leftmost bit to the rightmost position.

0x3C is 00111100. Rotating left by 1 bit gives 01111000 (0x78). Number of set bits is 4.