Bit manipulation techniques - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of bitwise AND operation

What is the output of the following C code?

C

int a = 12;  // binary: 1100
int b = 10;  // binary: 1010
int c = a & b;
printf("%d", c);

A 8

B 14

C 2

D 10

💡 Hint

Remember that & compares bits and returns 1 only if both bits are 1.

12 in binary is 1100 and 10 is 1010. Bitwise AND gives 1000 which is 8 in decimal.

❓ Predict Output

intermediate

Output of left shift operation

What will be printed by this C code?

C

unsigned int x = 5;  // binary: 0000 0101
unsigned int y = x << 2;
printf("%u", y);

A 10

B 2

C 20

D 40

💡 Hint

Left shift by 2 means multiply by 2 twice (4 times).

5 shifted left by 2 bits is 5 * 4 = 20.

❓ Predict Output

advanced

Output of toggling bits using XOR

What is the output of this C code snippet?

C

int num = 29;  // binary: 0001 1101
int mask = 15; // binary: 0000 1111
int result = num ^ mask;
printf("%d", result);

A 18

B 14

C 12

D 31

💡 Hint

XOR flips bits where mask has 1s.

29 (00011101) XOR 15 (00001111) flips last 4 bits: 00010010 which is 18 decimal.

❓ Predict Output

advanced

Output of checking if a number is a power of two

What does this C code print?

C

int n = 16;
int check = (n & (n - 1)) == 0;
printf("%d", check);

A 16

B 0

C 1

D 15

💡 Hint

A power of two has only one bit set.

For powers of two, n & (n-1) equals zero, so check is 1 (true).

❓ Predict Output

expert

Output of clearing the rightmost set bit

What is the output of this C code?

C

int x = 44;  // binary: 0010 1100
int y = x & (x - 1);
printf("%d", y);

A 44

B 40

C 12

D 32

💡 Hint

x & (x-1) clears the rightmost set bit.

44 in binary is 00101100. Clearing rightmost set bit gives 00101000 which is 40.

Practice

1. What does the bitwise AND operator & do when applied between two bits?

easy

A. Returns 1 if either bit is 1

B. Returns 0 always

C. Flips the bits

D. Returns 1 only if both bits are 1

Solution

1. Step 1: Understand bitwise AND operation

   The bitwise AND operator compares each bit of two numbers and returns 1 only if both bits are 1.

2. Step 2: Apply to bits

   For example, 1 & 1 = 1, but 1 & 0 = 0 and 0 & 0 = 0.

3. Final Answer:

   Returns 1 only if both bits are 1 -> Option D

4. Quick Check:

   AND = 1 if both bits 1 [OK]

Hint: AND returns 1 only when both bits are 1 [OK]

Common Mistakes:

• Confusing AND with OR operator
• Thinking AND flips bits
• Assuming AND always returns 1

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

easy

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

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

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

D. num = num << 3;

Solution

1. Step 1: Understand setting a bit

   To set a bit means to make it 1 without changing other bits. We use OR with a mask that has 1 at the target bit.

2. Step 2: Use bit shift and OR

   The mask is created by shifting 1 left by 3 (1 << 3). OR-ing num with this mask sets the 3rd bit.

3. Final Answer:

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

4. Quick Check:

   Set bit = OR with shifted 1 [OK]

Hint: Use OR with (1 << bit_position) to set a bit [OK]

Common Mistakes:

• Using AND instead of OR to set bits
• Using XOR which toggles bits
• Shifting num instead of 1

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

unsigned int x = 5;  // binary 0101
x = x ^ (1 << 2);
printf("%u", x);

medium

A. 4

B. 1

C. 5

D. 7

Solution

1. Step 1: Understand initial value and mask

   x = 5 in binary is 0101. The mask is (1 << 2) = 0100 (decimal 4).

2. Step 2: Apply XOR operation

   XOR toggles bits where mask has 1. 0101 ^ 0100 = 1101 (decimal 7). Let's do bitwise XOR:
   Bits (3 2 1 0):
   5 = 0 1 0 1
   Mask = 0 1 0 0
   XOR = 0 0 0 1 is incorrect.
   Actually, bits are:
   Bit 3: 0 ^ 0 = 0
   Bit 2: 1 ^ 1 = 0
   Bit 1: 0 ^ 0 = 0
   Bit 0: 1 ^ 0 = 1
   Result: 0001 (decimal 1) is incorrect for toggling bit 2.
   Wait, bit 2 is the third bit from right (0-indexed). So bits are:
   Bit 0: least significant bit
   Bit 1: next
   Bit 2: next
   So 5 (0101) bits:
   Bit 3: 0
   Bit 2: 1
   Bit 1: 0
   Bit 0: 1
   XOR with mask 0100:
   Bit 3: 0 ^ 0 = 0
   Bit 2: 1 ^ 1 = 0
   Bit 1: 0 ^ 0 = 0
   Bit 0: 1 ^ 0 = 1
   Result: 0001 (decimal 1). So output is 1.
   Therefore, correct answer is 1 -> 1.

3. Final Answer:

   1 -> Option B

4. Quick Check:

   XOR toggles bit 2: 5 ^ 4 = 1 [OK]

Hint: XOR toggles bits where mask is 1 [OK]

Common Mistakes:

• Miscounting bit positions (0-indexed from right)
• Confusing XOR with OR or AND
• Incorrect binary to decimal conversion

4. Identify the error in this C code snippet that tries to clear the 1st bit of num:

unsigned int num = 14;  // binary 1110
num = num & (1 << 1);
printf("%u", num);

medium

A. Mask should be inverted before AND

B. Should use OR instead of AND

C. Bit shift is incorrect, should be (1 << 0)

D. No error, code is correct

Solution

1. Step 1: Understand clearing a bit

   To clear a bit, we AND with a mask that has 0 at the target bit and 1s elsewhere.

2. Step 2: Analyze given mask

   The mask (1 << 1) is 00010 (binary), which has 1 only at bit 1. AND-ing num with this mask keeps only bit 1 and clears others, not what we want.

3. Step 3: Correct mask usage

   We should invert the mask: ~ (1 << 1), so bit 1 is 0 and others 1, then AND with num to clear bit 1.

4. Final Answer:

   Mask should be inverted before AND -> Option A

5. Quick Check:

   Clear bit = AND with inverted mask [OK]

Hint: Clear bit by AND with ~(1 << bit_position) [OK]

Common Mistakes:

• Using mask without inversion to clear bits
• Using OR instead of AND to clear bits
• Confusing bit positions

5. You want to check if the 4th bit (0-indexed) of an unsigned int flags is set (1). Which code snippet correctly does this and prints "Set" if true, otherwise "Not set"?

hard

A. if ((flags ^ (1 << 4)) == 1) printf("Set"); else printf("Not set");

B. if ((flags | (1 << 4)) == 0) printf("Set"); else printf("Not set");

C. if ((flags & (1 << 4)) != 0) printf("Set"); else printf("Not set");

D. if ((flags & (1 << 3)) != 0) printf("Set"); else printf("Not set");

Solution

1. Step 1: Understand checking a bit

   To check if a bit is set, AND the number with a mask having 1 at that bit. If result is not zero, bit is set.

2. Step 2: Analyze options

   if ((flags & (1 << 4)) != 0) printf("Set"); else printf("Not set"); uses (flags & (1 << 4)) != 0 which correctly checks the 4th bit.
   if ((flags | (1 << 4)) == 0) printf("Set"); else printf("Not set"); uses OR and compares to zero, which is incorrect.
   if ((flags ^ (1 << 4)) == 1) printf("Set"); else printf("Not set"); uses XOR and compares to 1, which is incorrect.
   if ((flags & (1 << 3)) != 0) printf("Set"); else printf("Not set"); checks bit 3 instead of 4, so wrong bit.

3. Final Answer:

   if ((flags & (1 << 4)) != 0) printf("Set"); else printf("Not set"); -> Option C

4. Quick Check:

   Check bit = AND with mask != 0 [OK]

Hint: Check bit with AND and compare to zero [OK]

Common Mistakes:

• Using OR or XOR to check bits
• Checking wrong bit position
• Comparing AND result incorrectly