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 initial integer value 8 (binary 1000), what is the output after setting the 3rd bit (0-based from right)?

DSA C

int x = 8; // binary 1000
int bit = 3;
x = x | (1 << bit);
printf("%d\n", x);

A 8

B 12

C 16

D 24

💡 Hint

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

Initial value 8 (binary 1000). Bit positions (0-based from right): bit 0=0, bit 1=0, bit 2=0, bit 3=1. Setting bit 3: 1 << 3 = 8 (binary 1000). 1000 | 1000 = 1000 (decimal 8).

❓ Predict Output

intermediate

What is the output after clearing the 2nd bit?

Given the integer value 15 (binary 1111), what is the output after clearing the 2nd bit (0-based from right)?

DSA C

int x = 15; // binary 1111
int bit = 2;
x = x & ~(1 << bit);
printf("%d\n", x);

A 7

B 15

C 11

D 8

💡 Hint

Clearing a bit means setting it to 0 using AND with the negation of the bit mask.

Clearing bit 2 means turning off the bit at position 2. 1 << 2 is 4 (binary 0100). Negation is 1011. 1111 & 1011 = 1011 (decimal 11).

❓ Predict Output

advanced

What is the output after toggling the 1st bit?

Given the integer value 10 (binary 1010), what is the output after toggling the 1st bit (0-based from right)?

DSA C

int x = 10; // binary 1010
int bit = 1;
x = x ^ (1 << bit);
printf("%d\n", x);

A 14

B 8

C 10

D 12

💡 Hint

Toggling a bit flips it: if it was 1, it becomes 0; if 0, it becomes 1.

Toggling bit 1 means XOR with 1 << 1 (2). 10 (1010) XOR 2 (0010) = 8 (1000).

🧠 Conceptual

advanced

Which operation clears a specific bit in an integer?

You want to clear (set to 0) the bit at position k in an integer x. Which of the following operations correctly does this?

A x = x | (1 << k);

B x = x & (1 << k);

C x = x ^ (1 << k);

D x = x & ~(1 << k);

💡 Hint

Clearing a bit means making sure that bit is 0 while keeping others unchanged.

Using AND with the negation of the bit mask clears the bit at position k.

🔧 Debug

expert

Why does this code fail to toggle the 4th bit?

Consider this code snippet intended to toggle the 4th bit of x: int x = 20; int bit = 4; x = x ^ 1 << bit; printf("%d\n", x); Why might this code not work as expected?

DSA C

int x = 20;
int bit = 4;
x = x ^ 1 << bit;
printf("%d\n", x);

A Operator precedence causes the XOR to apply before the shift, so the wrong bit is toggled.

B The bit index 4 is out of range for an int, causing undefined behavior.

C The code is missing parentheses around the shift operation, causing a syntax error.

D The variable x is not initialized before use.

💡 Hint

Check operator precedence between ^ and << in C.

In C, << has higher precedence than ^, so x ^ 1 << bit is evaluated as (x ^ 1) << bit, which is not intended. Parentheses are needed: x ^ (1 << bit).