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 ON (1) or OFF (0). You want to learn how to change these bits using basic bit operations.

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

📋 What You'll Learn

Create an integer variable with a specific value

Create a mask variable for bit operations

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

Print the results after 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 roles in embedded systems, systems programming, and software development that requires efficient data handling.

1

Create the initial number

Create an integer variable called number and set it to 29.

DSA C++

// Your code here

Hint

Think of 29 as the bits you want to change. 29 in binary is 00011101.

2

Create a mask for bit operations

Create an integer variable called mask and set it to 21.

DSA C++

int number = 29;
// Your code here

Hint

21 in binary is 00010101. This mask will help you change bits in number.

3

Apply bitwise AND, OR, XOR, NOT, left shift, and right shift

Create six integer variables: and_result, or_result, xor_result, not_result, left_shift, and right_shift. Set them to the results of applying bitwise AND, OR, XOR between number and mask, bitwise NOT of number, left shift number by 2, and right shift number by 2 respectively.

DSA C++

int number = 29;
int mask = 21;
// Your code here

Hint

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

4

Print the results

Print the values of and_result, or_result, xor_result, not_result, left_shift, and right_shift each on a new line.

DSA C++

int number = 29;
int mask = 21;
int and_result = number & mask;
int or_result = number | mask;
int xor_result = number ^ mask;
int not_result = ~number;
int left_shift = number > 2;

// Your code here

Hint

Use std::cout to print each result on its own line.

Practice

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

easy

A. 0

B. 1

C. 6

D. 7

Solution

1. Step 1: Convert numbers to binary

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

2. Step 2: Perform bitwise AND

   0101 AND 0011 = 0001 (only bits where both are 1 remain 1).

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
• Assuming bitwise AND adds numbers

2. Which of the following is the correct syntax for performing a left shift by 2 bits on an integer variable x 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++, left shift is done using << operator.

2. Step 2: Check syntax correctness

   x << 2; uses x << 2; which is correct syntax for left shift by 2 bits.

3. Final Answer:

   x << 2; -> Option A

4. Quick Check:

   Left shift uses << operator [OK]

Hint: Left shift uses <<, right shift uses >> [OK]

Common Mistakes:

• Using >>> which is not valid in C++
• Confusing left shift with right shift
• Adding extra symbols after operator

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

int x = 12;  // binary 1100
int y = 5;   // binary 0101
int z = x ^ y;
std::cout << z << std::endl;

medium

A. 8

B. 15

C. 9

D. 7

Solution

1. Step 1: Convert numbers to binary

   x = 12 (1100), y = 5 (0101).

2. Step 2: Perform bitwise XOR

   1100 XOR 0101 = 1001 (bits differ become 1).

3. Step 3: Convert result back to decimal

   1001 in decimal is 9.

4. Final Answer:

   9 -> Option C

5. Quick Check:

   12 ^ 5 = 9 [OK]

Hint: XOR sets bits where inputs differ [OK]

Common Mistakes:

• Mixing XOR with OR or AND
• Not converting binary to decimal correctly
• Assuming XOR adds numbers

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

int a = 8;
int b = a >> 2;
int c = ~b;
std::cout << c << std::endl;

medium

A. No error; code runs and prints the result

B. Using bitwise NOT (~) on an integer is invalid

C. Right shift operator (>>) is used incorrectly

D. Missing semicolon after variable declaration

Solution

1. Step 1: Check each operation

   Right shift (>>) on integer 8 by 2 bits is valid: 8 (1000) >> 2 = 2 (0010).

2. Step 2: Check bitwise NOT usage

   Bitwise NOT (~) on integer 2 flips bits; this is valid in C++.

3. Step 3: Verify syntax

   All semicolons are present; code compiles and runs.

4. Final Answer:

   No error; code runs and prints the result -> Option A

5. Quick Check:

   Bitwise NOT and shifts are valid [OK]

Hint: Bitwise NOT and shifts are valid on integers [OK]

Common Mistakes:

• Thinking ~ is invalid on integers
• Confusing shift direction
• Assuming missing semicolon without checking

5. Given an integer n = 29, which of the following expressions correctly clears (sets to 0) the 3rd bit (counting from 0, right to left) using bit manipulation?

hard

A. ~n & (1 << 3)

B. n | (1 << 3)

C. n ^ (1 << 3)

D. n & ~(1 << 3)

Solution

1. Step 1: Understand clearing a bit

   To clear a bit, use AND with NOT mask that has 1 at target bit.

2. Step 2: Analyze each option

   n & ~(1 << 3): n & ~(1 << 3) clears 3rd bit correctly.
   n | (1 << 3): n | (1 << 3) sets 3rd bit.
   n ^ (1 << 3): n ^ (1 << 3) toggles 3rd bit.
   ~n & (1 << 3): ~n & (1 << 3) isolates 3rd bit of inverted n, not clearing.

3. Final Answer:

   n & ~(1 << 3) -> Option D

4. Quick Check:

   Clear bit = AND with NOT mask [OK]

Hint: Clear bit: AND with NOT of bit mask [OK]

Common Mistakes:

• Using OR instead of AND to clear bits
• Confusing toggle (XOR) with clear
• Misunderstanding bit indexing