OR for setting bits in Embedded C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this bitwise OR operation?

Consider the following C code snippet that uses bitwise OR to set bits in a byte.

unsigned char flags = 0x12;  // binary: 00010010
flags = flags | 0x04;         // set bit 2
printf("0x%02X", flags);

What will be printed?

Embedded C

unsigned char flags = 0x12;
flags = flags | 0x04;
printf("0x%02X", flags);

A 0x10

B 0x16

C 0x14

D 0x04

💡 Hint

Bitwise OR sets bits that are set in either operand.

0x12 is binary 00010010. OR with 0x04 (00000100) sets bit 2, resulting in 00010110 which is 0x16.

❓ Predict Output

intermediate

What is the value of 'status' after setting bits?

Given this C code:

unsigned int status = 0x0F0F;
status |= 0x00F0;
printf("0x%04X", status);

What is the output?

Embedded C

unsigned int status = 0x0F0F;
status |= 0x00F0;
printf("0x%04X", status);

A 0x0FFF

B 0x0F0F

C 0x00F0

D 0x0FF0

💡 Hint

OR operation sets bits that are set in either operand.

0x0F0F OR 0x00F0 sets bits in the middle nibble, resulting in 0x0FFF.

🔧 Debug

advanced

Why does this code not set the intended bits?

Look at this code snippet:

unsigned char reg = 0x01;
reg = reg | 0x02;
reg = reg & 0xFD;
printf("0x%02X", reg);

The programmer wants to set bit 1 and clear bit 2. What is the output and why?

Embedded C

unsigned char reg = 0x01;
reg = reg | 0x02;
reg = reg & 0xFD;
printf("0x%02X", reg);

A 0x02; only bit 1 is set

B 0x01; bit 1 was never set

C 0x05; bits 0 and 2 are set

D 0x03; bit 2 is not cleared because 0xFD clears bit 1 instead

💡 Hint

Check which bit 0xFD clears in binary.

0xFD is 11111101 in binary, which clears bit 1 (second bit), not bit 2. So bit 1 is cleared after setting it, resulting in 0x01 (bits 0 set only).

📝 Syntax

advanced

Which option correctly sets bits 0 and 3 using OR?

Which of the following C statements correctly sets bits 0 and 3 of variable flags without changing other bits?

Embedded C

unsigned char flags = 0x00;

A flags |= (1 << 0) | (1 << 3);

B flags = flags | 1 << 0 | 1 << 3;

C flags |= 1 << (0 | 3);

D flags = flags | (1 << 0 & 1 << 3);

💡 Hint

Remember operator precedence and grouping with parentheses.

Option A uses parentheses to group the OR of shifted bits correctly. Option A lacks parentheses and shifts incorrectly due to precedence. Option A shifts by (0|3)=3 only. Option A uses AND instead of OR.

🚀 Application

expert

How many bits are set after this sequence of OR operations?

Given this code:

unsigned char mask = 0x00;
mask |= 1 << 1;
mask |= 1 << 4;
mask |= 1 << 7;

How many bits are set to 1 in mask after these operations?

Embedded C

unsigned char mask = 0x00;
mask |= 1 << 1;
mask |= 1 << 4;
mask |= 1 << 7;

A 1

B 2

C 3

D 4

💡 Hint

Count how many bits are set by each OR operation.

Each OR sets one distinct bit: bit 1, bit 4, and bit 7. Total bits set = 3.