Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to perform bitwise AND of two numbers.

DSA Go

result := a [1] b

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

A<<

B&

C^

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using '|' which is bitwise OR instead of AND.

Using '^' which is bitwise XOR instead of AND.

The '&' operator performs bitwise AND, which sets each bit to 1 only if both bits are 1.

2fill in blank

medium

Complete the code to perform bitwise XOR of two numbers.

DSA Go

result := x [1] y

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

A^

B|

C&

D>>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '&' which is bitwise AND instead of XOR.

Using '|' which is bitwise OR instead of XOR.

The '^' operator performs bitwise XOR, which sets each bit to 1 only if the bits are different.

3fill in blank

hard

Fix the error in the code to perform left shift by 3 bits.

DSA Go

shifted := num [1] 3

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

A<<

B&

C>>

D^

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>>' which is right shift instead of left shift.

Using '^' which is XOR instead of shift.

The '<<' operator shifts bits to the left, multiplying the number by 2 to the power of the shift count.

4fill in blank

hard

Fill both blanks to create a mask with 1s in the lowest 4 bits and clear other bits.

DSA Go

mask := (1 [1] 4) [2] 1

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

A<<

B-

C^

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' instead of '-' which gives wrong mask.

Using '^' which is XOR and does not create mask.

Shift 1 left by 4 bits to get 0b10000, then subtract 1 to get 0b1111 which is the mask for lowest 4 bits.

5fill in blank

hard

Fill all three blanks to toggle the 2nd bit of number n.

DSA Go

mask := 1 [1] 1
result := n [2] mask
final := ^result [3] mask

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

A<<

B^

C&

D|

Attempts:

3 left

💡 Hint

Common Mistakes

Using AND instead of OR to set bits.

Using right shift instead of left shift for mask.

Shift 1 left by 1 to get mask for 2nd bit, XOR toggles that bit in n, OR with mask sets that bit.

Practice

1. What is the result of the bitwise AND operation between 6 and 3? (6 = 110 in binary, 3 = 011 in binary)

easy

A. 2

B. 1

C. 3

D. 0

Solution

1. Step 1: Convert numbers to binary

   6 in binary is 110, 3 in binary is 011.

2. Step 2: Perform bitwise AND

   110 AND 011 = 010 in binary, which is 2 in decimal.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   6 & 3 = 2 [OK]

Hint: AND keeps bits set in both numbers only [OK]

Common Mistakes:

• Confusing AND with OR operation
• Not converting numbers to binary before operation
• Mixing decimal and binary values

2. Which of the following is the correct syntax in Go to perform a bitwise XOR between variables a and b?

easy

A. c := a & b

B. c := a && b

C. c := a ^ b

D. c := a | b

Solution

1. Step 1: Recall Go bitwise operators

   In Go, '^' is the bitwise XOR operator.

2. Step 2: Match syntax

   c := a ^ b uses '^' correctly for XOR. Other options use logical AND (&&), OR (|), and AND (&).

3. Final Answer:

   c := a ^ b -> Option C

4. Quick Check:

   Bitwise XOR uses '^' in Go [OK]

Hint: XOR operator in Go is '^' symbol [OK]

Common Mistakes:

• Using '&&' which is logical AND, not bitwise
• Confusing OR '|' with XOR '^'
• Using '&' which is bitwise AND, not XOR

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

package main
import "fmt"
func main() {
  var x uint8 = 12  // binary 00001100
  var y uint8 = 5   // binary 00000101
  fmt.Println(x | y)
  fmt.Println(x & y)
  fmt.Println(x ^ y)
}

medium

A. 15
4
9

B. 14
4
9

C. 15
5
8

D. 13
4
9

Solution

1. Step 1: Convert x and y to binary

   x = 12 = 00001100, y = 5 = 00000101.

2. Step 2: Calculate bitwise OR, AND, XOR

   OR: 00001100 | 00000101 = 00001101 (13 decimal)
   AND: 00001100 & 00000101 = 00000100 (4 decimal)
   XOR: 00001100 ^ 00000101 = 00001001 (9 decimal)

3. Step 3: Verify output

   Output lines are:
   13
   4
   9

4. Final Answer:

   13 4 9 -> Option D

5. Quick Check:

   Bitwise OR, AND, XOR results match 13, 4, 9 [OK]

Hint: OR sets bits from either number, AND only common bits [OK]

Common Mistakes:

• Mixing up OR and XOR results
• Incorrect binary conversion
• Confusing decimal output values

4. The following Go code is intended to left shift the number 3 by 2 bits. Identify the error.

package main
import "fmt"
func main() {
  var a int = 3
  var b int = 2
  var c int = a << b
  fmt.Println(c)
}

medium

A. Error: shift count must be unsigned integer

B. No error, output is 12

C. Error: cannot shift int by int

D. Error: missing parentheses around shift

Solution

1. Step 1: Understand Go shift rules

   In Go, shift count must be unsigned integer or untyped constant. Here, b is int but allowed.

2. Step 2: Calculate left shift

   3 << 2 means shifting bits of 3 (00000011) left by 2 bits: 00001100 = 12 decimal.

3. Final Answer:

   No error, output is 12 -> Option B

4. Quick Check:

   3 << 2 = 12 [OK]

Hint: Shift count can be int variable in Go if within range [OK]

Common Mistakes:

• Assuming shift count must be uint explicitly
• Thinking parentheses are required around shift
• Confusing shift operators with arithmetic operators

5. Given a 8-bit unsigned integer n, write a Go expression using bitwise operators to toggle (flip) the 3rd bit (counting from 0, right to left) of n. Which expression correctly does this?

hard

A. n ^ (1 << 3)

B. n & (1 << 3)

C. n | (1 << 3)

D. ~n & (1 << 3)

Solution

1. Step 1: Understand toggling a bit

   Toggling means flipping the bit: if 0 becomes 1, if 1 becomes 0.

2. Step 2: Use XOR to toggle

   XOR with 1 flips the bit, XOR with 0 keeps it same. So, n ^ (1 << 3) flips the 3rd bit.

3. Step 3: Check other options

   AND keeps bit only if both are 1 (no toggle), OR sets bit to 1 (no toggle), NOT and AND combination does not toggle.

4. Final Answer:

   n ^ (1 << 3) -> Option A

5. Quick Check:

   XOR with mask toggles bit [OK]

Hint: XOR with 1 flips bit, XOR with 0 keeps bit same [OK]

Common Mistakes:

• Using AND or OR which don't toggle bits
• Confusing NOT operator usage
• Incorrect bit position shift