AND for masking bits in Embedded C

Introduction

AND is used to keep certain bits of a number while turning others off. This helps us focus on specific parts of data.

You want to check if a specific light is on in a group of lights.

You need to get only the last 4 bits of a number to read a small value.

You want to clear some bits but keep others unchanged.

You want to test if a particular feature flag is active in a settings byte.

Syntax

Embedded C

result = value & mask;

The '&' symbol means AND in C language.

The mask is a number where bits you want to keep are 1, and bits to ignore are 0.

Examples

This keeps bits where mask has 1s and turns off others.

Embedded C

int value = 0b1101;  // binary 1101 (decimal 13)
int mask = 0b0101;   // binary 0101 (decimal 5)
int result = value & mask;  // result is 0b0101 (decimal 5)

This extracts the last 4 bits of value.

Embedded C

int value = 0xFF;    // hex FF (decimal 255)
int mask = 0x0F;     // hex 0F (decimal 15)
int result = value & mask;  // result is 0x0F (decimal 15)

Sample Program

This program uses AND to keep only the last 4 bits of a byte and prints the result in binary and decimal.

Embedded C

#include <stdio.h>

int main() {
    unsigned char value = 0b10101100;  // example byte
    unsigned char mask = 0b00001111;   // mask to keep last 4 bits
    unsigned char result = value & mask;

    printf("Value in binary: 0b10101100\n");
    printf("Mask in binary:  0b00001111\n");
    printf("Result in binary: 0b");

    for (int i = 7; i >= 0; i--) {
        printf("%d", (result >> i) & 1);
    }
    printf("\n");

    printf("Result in decimal: %d\n", result);
    return 0;
}

Important Notes

Mask bits set to 1 keep the original bit value; bits set to 0 turn off that bit.

AND masking does not change the original value; it creates a new result.

Use binary or hex notation to clearly see which bits are set in the mask.

Summary

AND with a mask keeps only the bits you want and turns off others.

Use 1s in the mask where you want to keep bits, 0s where you want to ignore.

This is useful to extract or check parts of data stored in bits.

Practice

1. What does the bitwise AND operation with a mask do in embedded C?

easy

A. It adds the mask value to the original number.

B. It keeps only the bits where the mask has 1s and turns off others.

C. It flips all bits in the original number.

D. It shifts the bits to the left by the mask value.

Solution

1. Step 1: Understand bitwise AND with mask

   The AND operation compares each bit of two numbers and returns 1 only if both bits are 1.

2. Step 2: Apply mask logic

   Using a mask with 1s keeps those bits from the original number; 0s turn off bits.

3. Final Answer:

   It keeps only the bits where the mask has 1s and turns off others. -> Option B

4. Quick Check:

   AND with mask = keep bits [OK]

Hint: Mask with 1 keeps bits; 0 clears bits [OK]

Common Mistakes:

• Thinking AND flips bits
• Confusing AND with addition
• Assuming AND shifts bits

2. Which of the following is the correct syntax to mask the lower 4 bits of a variable val in embedded C?

easy

A. val & 0xF

B. val | 0xF

C. val << 4

D. val ^ 0xF

Solution

1. Step 1: Identify mask for lower 4 bits

   Lower 4 bits mask is 0xF (binary 00001111).

2. Step 2: Apply bitwise AND syntax

   Use val & 0xF to keep only lower 4 bits.

3. Final Answer:

   val & 0xF -> Option A

4. Quick Check:

   AND with 0xF masks lower 4 bits [OK]

Hint: Use & with 0xF to keep lower 4 bits [OK]

Common Mistakes:

• Using OR instead of AND
• Using XOR which toggles bits
• Using shift instead of mask

3. What is the output of the following code?

unsigned char val = 0xAB;  // 10101011 in binary
unsigned char mask = 0x0F; // 00001111 in binary
unsigned char result = val & mask;
printf("%02X", result);

medium

A. 0B

B. AF

C. 0A

D. AB

Solution

1. Step 1: Convert values to binary

   val = 10101011, mask = 00001111.

2. Step 2: Perform bitwise AND

   10101011 & 00001111 = 00001011 (binary) = 0x0B (hex).

3. Final Answer:

   0B -> Option A

4. Quick Check:

   AND keeps lower 4 bits = 0B [OK]

Hint: AND keeps bits where mask is 1 [OK]

Common Mistakes:

• Reading hex incorrectly
• Confusing mask bits
• Assuming output is original val

4. The following code is intended to mask the upper 4 bits of val, but it does not work as expected. What is the error?

unsigned char val = 0xAB;  // 10101011
unsigned char mask = 0xF0; // 11110000
unsigned char result = val | mask;
printf("%X", result);

medium

A. Mask value is incorrect

B. Missing parentheses around expression

C. Variable types are wrong

D. Using OR (|) instead of AND (&) for masking

Solution

1. Step 1: Understand masking operation

   Masking to keep bits requires AND (&), not OR (|).

2. Step 2: Identify incorrect operator

   The code uses OR which sets bits instead of masking.

3. Final Answer:

   Using OR (|) instead of AND (&) for masking -> Option D

4. Quick Check:

   Masking needs AND, not OR [OK]

Hint: Mask bits with & not | [OK]

Common Mistakes:

• Using OR instead of AND
• Changing mask value unnecessarily
• Assuming parentheses fix bitwise errors

5. You have a 16-bit sensor data stored in uint16_t data. The upper 8 bits represent temperature, and the lower 8 bits represent humidity. How do you extract only the temperature value using AND masking?

hard

A. temperature = (data | 0xFF00) >> 8;

B. temperature = data & 0x00FF;

C. temperature = (data & 0xFF00) >> 8;

D. temperature = data >> 8;

Solution

1. Step 1: Identify mask for upper 8 bits

   Upper 8 bits mask is 0xFF00 (binary 11111111 00000000).

2. Step 2: Apply AND and shift right

   Use (data & 0xFF00) to keep upper bits, then shift right 8 to get value.

3. Final Answer:

   temperature = (data & 0xFF00) >> 8; -> Option C

4. Quick Check:

   Mask upper 8 bits and shift right [OK]

Hint: Mask upper bits with 0xFF00 then shift right 8 [OK]

Common Mistakes:

• Masking lower bits instead
• Using OR instead of AND
• Forgetting to shift after masking