Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA C - Build from Scratch

Bit Manipulation Basics AND OR XOR NOT Left Right Shift

📖 Scenario: Imagine you are working with a simple device that stores settings as bits in a number. Each bit can be turned on or off to control different features. You want to learn how to change these bits using basic bit operations.

🎯 Goal: You will create a small C program that uses bit manipulation operators: AND, OR, XOR, NOT, left shift, and right shift. You will see how these operators change the bits of numbers step-by-step.

📋 What You'll Learn

Create two integer variables with exact values

Create a variable to hold the result of bit operations

Use bitwise AND, OR, XOR, NOT, left shift, and right shift operators

Print the results of each operation

💡 Why This Matters

🌍 Real World

Bit manipulation is used in low-level programming, device control, and performance optimization where direct control of bits is needed.

💼 Career

Understanding bitwise operations is important for embedded systems developers, systems programmers, and anyone working with hardware or performance-critical code.

1

Create two integer variables

Create two integer variables called a and b with values 12 and 5 respectively.

DSA C

# Create two integer variables a and b with values 12 and 5
// Your code here

Hint

Use int a = 12; and int b = 5; to create the variables.

2

Create a variable for results

Create an integer variable called result and set it to 0.

DSA C

int a = 12;
int b = 5;
// Create an integer variable result and set it to 0
// Your code here

Hint

Use int result = 0; to create the variable.

3

Apply bitwise operations

Use the variables a and b to perform these bitwise operations and store each result in result: AND, OR, XOR, NOT of a, left shift a by 2, and right shift b by 1. Write one line for each operation.

DSA C

int a = 12;
int b = 5;
int result = 0;
// Perform bitwise AND of a and b and store in result
// Perform bitwise OR of a and b and store in result
// Perform bitwise XOR of a and b and store in result
// Perform bitwise NOT of a and store in result
// Perform left shift of a by 2 and store in result
// Perform right shift of b by 1 and store in result
// Your code here

Hint

Use & for AND, | for OR, ^ for XOR, ~ for NOT, << for left shift, and >> for right shift.

4

Print the results of each operation

Print the results of each bitwise operation in order: AND, OR, XOR, NOT of a, left shift a by 2, and right shift b by 1. Use printf with the format "%d\n" for each result.

DSA C

int a = 12;
int b = 5;
int result = 0;

result = a & b;  // AND
result = a | b;  // OR
result = a ^ b;  // XOR
result = ~a;     // NOT
result = a > 1; // Right shift

// Print the results of each operation below
// Your code here

Hint

Use printf("%d\n", result); after each operation to print the result on a new line.

Practice

1. What is the result of the bitwise AND operation between 5 and 3? (5 = 0101, 3 = 0011 in binary)

easy

A. 0

B. 1

C. 6

D. 7

Solution

1. Step 1: Convert numbers to binary

   5 in binary is 0101, and 3 in binary is 0011.

2. Step 2: Perform bitwise AND operation

   0101 AND 0011 = 0001 (only bits set in both numbers remain set).

3. Final Answer:

   1 -> Option B

4. Quick Check:

   5 & 3 = 1 [OK]

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

Common Mistakes:

• Confusing AND with OR operation
• Not converting decimal to binary correctly
• Mixing bit positions during calculation

2. Which of the following is the correct syntax for left shifting the integer x by 2 bits in C?

easy

A. x >> 2;

B. x <<< 2;

C. x << 2;

D. x >> 2 <<;

Solution

1. Step 1: Understand left shift operator in C

   The left shift operator is written as << and shifts bits to the left.

2. Step 2: Check syntax correctness

   x << 2; uses x << 2; which correctly shifts bits of x left by 2 positions.

3. Final Answer:

   x << 2; -> Option C

4. Quick Check:

   Left shift syntax = x << 2; [OK]

Hint: Use << for left shift, >> for right shift [OK]

Common Mistakes:

• Using >>> which is invalid in C
• Confusing left shift with right shift
• Incorrect operator placement or syntax

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

int x = 6; // binary 0110
int y = x ^ 3; // 3 = 0011
printf("%d", y);

medium

A. 9

B. 7

C. 1

D. 5

Solution

1. Step 1: Convert numbers to binary and apply XOR

   6 in binary is 0110, 3 is 0011. XOR operation: 0110 ^ 0011 = 0101.

2. Step 2: Calculate decimal value of XOR result

   0101 in binary equals 5 decimal.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   6 ^ 3 = 5 [OK]

Hint: XOR flips bits where bits differ [OK]

Common Mistakes:

• Mixing XOR with OR or AND
• Incorrect binary conversion
• Misreading bit positions

4. Identify the error in the following C code snippet:

int x = 8;
int y = ~x << 1;
printf("%d", y);

medium

A. Using ~ operator without parentheses causes wrong shift

B. Missing semicolon after declaration

C. Shift operator << cannot be used with ~ operator

D. No error, code runs correctly

Solution

1. Step 1: Understand operator precedence

   The bitwise NOT (~) has higher precedence than shift (<<), so ~x is evaluated first.

2. Step 2: Check if parentheses needed

   Without parentheses, ~x is computed, then shifted left. This is correct but may cause confusion if intended differently.

3. Step 3: Identify common mistake

   If intention was to shift x first then NOT, parentheses are needed. Otherwise, no syntax error but logic may be wrong.

4. Final Answer:

   Using ~ operator without parentheses causes wrong shift -> Option A

5. Quick Check:

   Operator precedence matters; use parentheses [OK]

Hint: Use parentheses to control bitwise operation order [OK]

Common Mistakes:

• Ignoring operator precedence
• Assuming shift applies before NOT
• Thinking code has syntax error when it doesn't

5. Given an integer n, which expression correctly checks if the 3rd bit (from right, zero-based) is set (1)?

hard

A. (n & (1 << 3)) != 0

B. (n | (1 << 3)) == 0

C. (n ^ (1 << 3)) == 0

D. (n >> 3) & 1 == 0

Solution

1. Step 1: Understand bit position and mask

   The 3rd bit (zero-based) means bit at position 3. Mask is 1 shifted left by 3: (1 << 3) = 8.

2. Step 2: Check if bit is set using AND

   Using (n & (1 << 3)) checks if that bit is 1. If result is not zero, bit is set.

3. Final Answer:

   (n & (1 << 3)) != 0 -> Option A

4. Quick Check:

   Bit check uses AND with mask [OK]

Hint: Use AND with (1 << bit_position) to check bit [OK]

Common Mistakes:

• Using OR or XOR instead of AND for checking
• Wrong bit position shift
• Comparing shifted bit to zero incorrectly