Set Clear Toggle a Specific Bit in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set the 3rd bit of number n.

DSA C++

n = n | (1 << [1]);

Drag options to blanks, or click blank then click option'

A1

B3

C0

D2

Attempts:

3 left

💡 Hint

Common Mistakes

Using 3 instead of 2 for the bit index.

Confusing bit positions starting at 1 instead of 0.

Bits are zero-indexed from the right. The 3rd bit means index 2, so shifting 1 by 2 sets the 3rd bit.

2fill in blank

medium

Complete the code to clear the 4th bit of number n.

DSA C++

n = n & ~(1 << [1]);

Drag options to blanks, or click blank then click option'

A4

B3

C2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 4 instead of 3 for the bit index.

Forgetting to invert the mask with ~.

Clearing the 4th bit means zeroing bit at index 3 (zero-based). So shift 1 by 3 and invert.

3fill in blank

hard

Fix the error in the code to toggle the 5th bit of number n.

DSA C++

n = n [1] (1 << 4);

Drag options to blanks, or click blank then click option'

A&

B|

C^

D~

Attempts:

3 left

💡 Hint

Common Mistakes

Using & or | which clear or set bits instead of toggling.

Using ~ which inverts all bits.

Toggling a bit uses XOR (^). XOR flips the bit if it is 1, else sets it if 0.

4fill in blank

hard

Fill both blanks to check if the 2nd bit of n is set and print "Yes" if true.

DSA C++

if (n [1] (1 << [2])) {
    std::cout << "Yes" << std::endl;
}

Drag options to blanks, or click blank then click option'

A&

B1

C^

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using | or ^ which do not check bits.

Using wrong bit index.

To check if a bit is set, use bitwise AND (&) with 1 shifted by the bit index. The 2nd bit is index 1.

5fill in blank

hard

Fill all three blanks to set the 1st bit, clear the 3rd bit of n.

DSA C++

n = n [1] (1 << 0);
n = n [2] ~(1 << [3]);

Drag options to blanks, or click blank then click option'

A|

B&

C2

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing operators for set and clear.

Using wrong bit index for clearing.

Set 1st bit with OR (|) and shift 0. Clear 3rd bit with AND (&) and inverted mask shifted by 2.

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