Bitwise AND, OR, XOR in C

Introduction

Bitwise AND, OR, and XOR let you work directly with the bits inside numbers. This helps you do fast, low-level operations like turning bits on or off.

Checking if a specific bit in a number is on or off.

Setting or clearing a particular bit in a number.

Combining two numbers bit by bit to create a new number.

Performing simple encryption or toggling bits.

Optimizing performance in low-level programming or embedded systems.

Syntax

C

/* Bitwise operations on integers */
int bitwise_and = a & b;  // AND operation
int bitwise_or = a | b;   // OR operation
int bitwise_xor = a ^ b;  // XOR operation

The operators work on each bit of the numbers separately.

Operands must be integers (int, char, etc.).

Examples

Basic example showing bitwise AND, OR, XOR on two numbers.

C

int a = 6;  // binary 0110
int b = 3;  // binary 0011
int and_result = a & b;  // 2 (0010)
int or_result = a | b;   // 7 (0111)
int xor_result = a ^ b;  // 5 (0101)

Edge case: Using zero and one to see how bits combine.

C

int zero = 0;
int one = 1;
int and_zero = zero & one;  // 0
int or_zero = zero | one;   // 1
int xor_zero = zero ^ one;  // 1

Using all bits set to 1 to see how AND, OR, XOR behave.

C

int all_ones = ~0;  // all bits set to 1
int a = 5;  // 0101
int and_all_ones = a & all_ones;  // 5 (unchanged)
int or_all_ones = a | all_ones;    // -1 (all bits 1)
int xor_all_ones = a ^ all_ones;   // bitwise NOT of a

Sample Program

This program shows how bitwise AND, OR, and XOR work on two numbers. It prints the results so you can see the effect on bits.

C

#include <stdio.h>

int main() {
    int number1 = 12;  // binary 1100
    int number2 = 10;  // binary 1010

    printf("Number1: %d (binary 1100)\n", number1);
    printf("Number2: %d (binary 1010)\n", number2);

    int and_result = number1 & number2;
    int or_result = number1 | number2;
    int xor_result = number1 ^ number2;

    printf("Bitwise AND (number1 & number2): %d\n", and_result);
    printf("Bitwise OR (number1 | number2): %d\n", or_result);
    printf("Bitwise XOR (number1 ^ number2): %d\n", xor_result);

    return 0;
}

Important Notes

Time complexity is O(1) because bitwise operations happen directly on CPU registers.

Space complexity is O(1) as no extra memory is needed beyond variables.

Common mistake: Confusing bitwise operators (&, |, ^) with logical operators (&&, ||, !).

Use bitwise operations when you need to manipulate bits directly, not for general true/false logic.

Summary

Bitwise AND (&), OR (|), and XOR (^) work on each bit of integers.

They are useful for checking, setting, or toggling bits efficiently.

Remember: bitwise operators are different from logical operators.

Practice

1. Which bitwise operator in C returns 1 only if both corresponding bits of two numbers are 1?
a & b means:

easy

A. Bitwise AND

B. Bitwise OR

C. Bitwise XOR

D. Logical AND

Solution

1. Step 1: Understand bitwise AND operation

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

2. Step 2: Compare with other operators

   Bitwise OR returns 1 if either bit is 1, XOR returns 1 if bits differ, and logical AND is different (works on boolean values).

3. Final Answer:

   Bitwise AND -> Option A

4. Quick Check:

   AND = & = Bitwise AND [OK]

Hint: AND means both bits must be 1 to get 1 [OK]

Common Mistakes:

• Confusing bitwise AND with logical AND (&&)
• Mixing AND with OR or XOR operators
• Thinking bitwise AND returns 1 if either bit is 1

2. Which of the following is the correct syntax to perform bitwise XOR on two integers a and b in C?

easy

A. a ^ b

B. a && b

C. a | b

D. a & b

Solution

1. Step 1: Identify bitwise XOR operator

   In C, the caret symbol (^) is used for bitwise XOR operation.

2. Step 2: Differentiate from other operators

   && is logical AND, | is bitwise OR, & is bitwise AND, so only ^ is XOR.

3. Final Answer:

   a ^ b -> Option A

4. Quick Check:

   ^ means XOR in C [OK]

Hint: XOR uses ^ symbol in C [OK]

Common Mistakes:

• Using && instead of ^ for XOR
• Confusing | (OR) with ^ (XOR)
• Using & (AND) instead of ^ (XOR)

3. What is the output of the following C code?

int a = 6;  // binary 0110
int b = 11; // binary 1011
printf("%d", a & b);

medium

A. 3

B. 15

C. 7

D. 2

Solution

1. Step 1: Convert numbers to binary

   6 in binary is 0110, 11 in binary is 1011.

2. Step 2: Perform bitwise AND

   0110 & 1011 = 0010 (only bits where both are 1 remain 1).

3. Step 3: Convert result back to decimal

   0010 in decimal is 2.

4. Final Answer:

   2 -> Option D

5. Quick Check:

   6 & 11 = 2 [OK]

Hint: AND keeps bits set in both numbers only [OK]

Common Mistakes:

• Confusing AND with OR or XOR results
• Not converting binary back to decimal correctly
• Mixing up bit positions

4. Identify the error in this C code snippet that tries to perform bitwise OR:

int a = 5;
int b = 9;
int c = a | b;
printf("%d", c);

medium

A. Missing semicolon after variable declarations

B. Incorrect operator used for bitwise OR

C. No error, code will print correct output

D. printf format specifier should be %x for bitwise OR

Solution

1. Step 1: Check syntax correctness

   All lines have semicolons; operator | is correct for bitwise OR.

2. Step 2: Verify printf usage

   %d is correct to print integer decimal value; no error present.

3. Final Answer:

   No error, code will print correct output -> Option C

4. Quick Check:

   Bitwise OR uses | and %d prints decimal [OK]

Hint: Bitwise OR uses | and %d prints decimal [OK]

Common Mistakes:

• Thinking | is wrong operator for OR
• Believing %x is mandatory for bitwise results
• Missing semicolons (not the case here)

5. Given two integers x = 12 and y = 10, which expression toggles the bits of x where y has bits set, using bitwise XOR?

hard

A. x | y

B. x ^ y

C. x & y

D. ~x ^ y

Solution

1. Step 1: Understand toggling bits with XOR

   Bitwise XOR toggles bits where the second operand has 1s.

2. Step 2: Analyze each option

   x & y keeps bits set in both; x | y sets bits set in either; ~x ^ y is unrelated toggling with complement.

3. Final Answer:

   x ^ y -> Option B

4. Quick Check:

   XOR toggles bits where second operand has 1 [OK]

Hint: XOR toggles bits where second number has 1s [OK]

Common Mistakes:

• Using AND or OR instead of XOR for toggling
• Confusing complement operator ~ with XOR
• Not understanding toggling effect of XOR