Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bitwise AND and OR Operations

What is the output of the following C++ code snippet?

DSA C++

int a = 12;  // binary: 1100
int b = 10;  // binary: 1010
int c = a & b;
int d = a | b;
std::cout << c << " " << d << std::endl;

A 8 14

B 4 14

C 8 12

D 4 12

💡 Hint

Recall that & compares bits and returns 1 only if both bits are 1; | returns 1 if either bit is 1.

12 in binary is 1100, 10 is 1010.
AND: 1100 & 1010 = 1000 (8 decimal)
OR: 1100 | 1010 = 1110 (14 decimal)

❓ Predict Output

intermediate

Result of XOR and NOT Operations

What will be printed by this C++ code?

DSA C++

unsigned int x = 5;  // binary: 0000 0101
unsigned int y = 9;  // binary: 0000 1001
unsigned int z = x ^ y;
unsigned int w = ~x;
std::cout << z << " " << w << std::endl;

A 12 4294967290

B 12 250

C 13 4294967290

D 13 250

💡 Hint

XOR returns 1 if bits differ; NOT flips all bits. Remember unsigned int is 32-bit.

5 (0000 0101) XOR 9 (0000 1001) = 0000 1100 (12 decimal).
NOT 5 flips bits: ~0000 0101 = 1111 1010 (in 32-bit unsigned, 4294967290).

❓ Predict Output

advanced

Left and Right Shift Effects on Signed Integers

What is the output of this C++ code snippet?

DSA C++

int a = -8;  // binary (two's complement): 1111 1000
int b = a >> 2;
int c = a << 2;
std::cout << b << " " << c << std::endl;

A 2 32

B -2 32

C -2 -32

D 2 -32

💡 Hint

Right shift on signed int usually preserves sign (arithmetic shift). Left shift multiplies by powers of two.

-8 >> 2 shifts bits right with sign extension: -8 / 4 = -2.
-8 << 2 shifts bits left: -8 * 4 = -32.

🧠 Conceptual

advanced

Understanding Bitwise NOT on Signed Integers

What is the decimal value of ~(-1) in a 32-bit signed integer system?

A -1

B 1

C 4294967294

D 0

💡 Hint

Remember that -1 in two's complement is all bits set to 1.

-1 in 32-bit two's complement is 0xFFFFFFFF.
Applying NOT flips all bits to 0x00000000 which is 0.

❓ Predict Output

expert

Complex Bitwise Expression Evaluation

What is the output of this C++ code?

DSA C++

unsigned int x = 29;  // binary: 0001 1101
unsigned int y = 15;  // binary: 0000 1111
unsigned int z = ((x & y) << 2) ^ (~y >> 1);
std::cout << z << std::endl;

A 4294967274

B 2147483596

C 4294967282

D 4294967266

💡 Hint

Break down the expression step-by-step: AND, shifts, NOT, then XOR.

x & y = 29 & 15 = 13 (0000 1101)
(x & y) << 2 = 13 << 2 = 52 (0011 0100)
~y = ~15 = 0xFFFFFFF0 (4294967280 unsigned)
~y >> 1 = 0x7FFFFFF8 (2147483640)
52 ^ 2147483640 = 2147483596

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