Bit manipulation techniques - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to set the 3rd bit of the variable num.

C

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

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

A3

B0

C2

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 3 instead of 2 for the bit position.

Confusing bit positions starting from 1 instead of 0.

Bits are counted from 0, so the 3rd bit is at position 2. Shifting 1 by 2 sets the 3rd bit.

2fill in blank

medium

Complete the code to clear (set to 0) the 5th bit of num.

C

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

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

A4

B5

C3

D6

Attempts:

3 left

💡 Hint

Common Mistakes

Using 5 instead of 4 for the bit position.

Forgetting to negate the mask before AND operation.

The 5th bit is at position 4 (counting from 0). We shift 1 by 4 and then invert it to clear that bit.

3fill in blank

hard

Fix the error in the code to toggle the 1st bit of num.

C

num = num [1] (1 << 0);

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

A|

B^

C&

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using AND or OR instead of XOR for toggling.

Using right shift operator which does not toggle bits.

The XOR operator (^) toggles bits. Using it with 1 shifted by 0 toggles the 1st bit.

4fill in blank

hard

Fill both blanks to check if the 4th bit of num is set (1).

C

if ((num [1] (1 << [2])) != 0) {
    // bit is set
}

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

A&

B|

C3

D4

Attempts:

3 left

💡 Hint

Common Mistakes

Using OR instead of AND for checking bits.

Using wrong bit position like 3 instead of 4.

Use bitwise AND (&) with 1 shifted by 4 to check if the 4th bit is set.

5fill in blank

hard

Fill all three blanks to create a mask that keeps only the lower 3 bits of num.

C

mask = (1 << [1]) [2] 1;
result = num [3] mask;

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

A3

B-

C&

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using OR instead of AND to apply the mask.

Using addition instead of subtraction in mask creation.

Using wrong bit count for shifting.

Shift 1 by 3, subtract 1 to get 0b0111 mask. Use AND to keep lower 3 bits.

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