Set Clear Toggle a Specific Bit in DSA C++ - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output after setting the 3rd bit?

Given the code below, what will be the printed value of num after setting the 3rd bit (0-based index from right)?

DSA C++

int num = 8; // binary 1000
int pos = 3;
num = num | (1 << pos);
std::cout << num << std::endl;

A 24

B 16

C 8

D 0

💡 Hint

Remember that bits are counted from 0 starting at the rightmost bit.

The 3rd bit (0-based) corresponds to the 8's place. Since num is already 8 (1000), setting the 3rd bit again keeps it 8.

❓ Predict Output

intermediate

What is the output after clearing the 1st bit?

What will be the output after clearing the 1st bit (0-based) of num?

DSA C++

int num = 11; // binary 1011
int pos = 1;
num = num & ~(1 << pos);
std::cout << num << std::endl;

A 3

B 10

C 11

D 9

💡 Hint

Clearing a bit means setting it to 0.

The 1st bit of 1011 is 1. Clearing it changes 1011 to 1001 which is 9.

❓ Predict Output

advanced

What is the output after toggling the 2nd bit?

Given the code below, what will be the output after toggling the 2nd bit (0-based) of num?

DSA C++

int num = 13; // binary 1101
int pos = 2;
num = num ^ (1 << pos);
std::cout << num << std::endl;

A 9

B 13

C 15

D 5

💡 Hint

Toggling flips the bit: 0 becomes 1, 1 becomes 0.

The 2nd bit of 1101 is 1. Toggling it changes 1101 to 1001 which is 9.

🔧 Debug

advanced

Identify the error in clearing a bit code

The following code is intended to clear the 4th bit of num. What error will it cause?

DSA C++

int num = 31; // binary 11111
int pos = 4;
num = num & (1 << ~pos);
std::cout << num << std::endl;

A Wrong output: 15

B SyntaxError

C Wrong output: 16

D Runtime error

💡 Hint

Check the operator precedence and bitwise NOT usage.

The expression (1 << ~pos) shifts 1 by a negative number (since ~4 = -5), which is undefined behavior and results in wrong output 15 instead of clearing bit 4.

🧠 Conceptual

expert

How many bits are set after toggling bits 0, 1, and 2 of 0?

Starting with num = 0, you toggle bits at positions 0, 1, and 2 (0-based). How many bits are set in num after these operations?

DSA C++

int num = 0;
num = num ^ (1 << 0);
num = num ^ (1 << 1);
num = num ^ (1 << 2);

A 1

B 3

C 2

D 0

💡 Hint

Toggling a bit from 0 sets it to 1.

Toggling bits 0,1,2 from 0 sets all three bits to 1, so 3 bits are set.

Practice

1. What does the operation num |= (1 << k) do to the integer num in C++?

easy

A. Toggles the k-th bit of num

B. Sets the k-th bit of num to 1

C. Clears the k-th bit of num to 0

D. Checks if the k-th bit of num is set

Solution

1. Step 1: Understand the bit shift operation

   The expression 1 << k creates a number with only the k-th bit set to 1 and all others 0.

2. Step 2: Apply bitwise OR to set the bit

   Using |= with this mask sets the k-th bit of num to 1 without changing other bits.

3. Final Answer:

   Sets the k-th bit of num to 1 -> Option B

4. Quick Check:

   num |= (1 << k) sets bit k [OK]

Hint: Use OR with shifted 1 to set a bit [OK]

Common Mistakes:

• Confusing set with clear operation
• Using AND instead of OR to set bits
• Misunderstanding bit positions starting from 0

2. Which of the following is the correct syntax to clear the 3rd bit (0-indexed) of an integer variable num in C++?

easy

A. num &= ~(1 << 3);

B. num |= (1 << 3);

C. num ^= (1 << 3);

D. num = (num & (1 << 3));

Solution

1. Step 1: Create mask to clear bit

   1 << 3 sets only the 3rd bit. Negating it with ~ creates a mask with all bits 1 except the 3rd bit 0.

2. Step 2: Use AND assignment to clear bit

   Applying &= with this mask clears the 3rd bit of num while keeping others unchanged.

3. Final Answer:

   num &= ~(1 << 3); -> Option A

4. Quick Check:

   AND with negated mask clears bit [OK]

Hint: Use AND with negated shifted 1 to clear bit [OK]

Common Mistakes:

• Using OR instead of AND to clear bits
• Forgetting to negate the mask
• Using XOR which toggles instead of clears

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

int num = 10; // binary 1010
num ^= (1 << 1);
std::cout << num << std::endl;

medium

A. 8

B. 10

C. 12

D. 14

Solution

1. Step 1: Understand initial value and bit positions

   num = 10 in binary is 1010. Bits are indexed from 0 (rightmost). Bit 1 is currently 1.

2. Step 2: Toggle the 1st bit using XOR

   num ^= (1 << 1) toggles bit 1. Since bit 1 is 1, toggling it changes it to 0. New binary: 1000 (decimal 8).

3. Final Answer:

   8 -> Option A

4. Quick Check:

   Toggle bit 1 of 1010 = 1000 = 8 [OK]

Hint: XOR toggles bits; 1 flips, 0 stays same [OK]

Common Mistakes:

• Confusing toggle with set or clear
• Miscounting bit positions starting at 1 instead of 0
• Assuming XOR always sets bit

4. Identify the error in the following code snippet that attempts to clear the 2nd bit of num:

int num = 15;
num = num & (1 << 2);
std::cout << num << std::endl;

medium

A. It sets only the 2nd bit instead of clearing it

B. It toggles the 2nd bit instead of clearing it

C. It clears all bits except the 2nd bit

D. It causes a syntax error due to missing parentheses

Solution

1. Step 1: Analyze the mask used

   The mask (1 << 2) has only the 2nd bit set to 1, all others 0.

2. Step 2: Understand the effect of AND with mask

   ANDing num with this mask keeps only the 2nd bit of num and clears all others. This is opposite of clearing the 2nd bit.

3. Final Answer:

   It clears all bits except the 2nd bit -> Option C

4. Quick Check:

   AND with mask keeps bit, does not clear it [OK]

Hint: Use AND with negated mask to clear bits, not plain mask [OK]

Common Mistakes:

• Using mask directly instead of negated mask
• Confusing AND with OR for clearing bits
• Assuming AND with mask clears bit

5. Given an integer num, which expression correctly toggles the 0th bit only if it is currently set (1), otherwise leaves num unchanged?

hard

A. num ^= 1;

B. num &= ~1;

C. num |= 1;

D. if ((num & 1) == 1) num ^= 1;

Solution

1. Step 1: Check if 0th bit is set

   The condition (num & 1) == 1 checks if the least significant bit is 1.

2. Step 2: Toggle 0th bit only if set

   If the bit is set, num ^= 1 toggles it to 0. Otherwise, no change is made.

3. Final Answer:

   if ((num & 1) == 1) num ^= 1; -> Option D

4. Quick Check:

   Toggle only if bit set using condition [OK]

Hint: Check bit before toggling to avoid unwanted changes [OK]

Common Mistakes:

• Toggling bit unconditionally
• Using OR or AND which always set or clear bit
• Not checking bit state before toggle