Bit Manipulation Basics AND OR XOR NOT Left Right Shift in DSA Go - Build from Scratch

Bit Manipulation Basics AND OR XOR NOT Left Right Shift

📖 Scenario: Imagine you are working with a simple device that stores settings as bits in a number. Each bit can be ON (1) or OFF (0). You want to learn how to change these bits using basic bit operations.

🎯 Goal: You will create a program that uses bit manipulation operations: AND, OR, XOR, NOT, left shift, and right shift on given numbers. You will see how these operations change the bits step-by-step.

📋 What You'll Learn

Create two variables with exact values

Create a variable for a shift amount

Use bitwise AND, OR, XOR, NOT, left shift, and right shift operations

Print the results of each operation

💡 Why This Matters

🌍 Real World

Bit manipulation is used in low-level programming, device control, and performance optimization where direct control of bits is needed.

💼 Career

Understanding bitwise operations is important for embedded systems, network programming, cryptography, and systems programming roles.

1

Create two integer variables

Create two integer variables called a and b with values 12 and 5 respectively.

DSA Go

# Create two integer variables a and b with values 12 and 5
// Your code here

Hint

Use := to create variables a and b with the exact values.

2

Create a shift amount variable

Create an integer variable called shift and set it to 2.

DSA Go

package main

import "fmt"

func main() {
    a := 12
    b := 5
    // Create a variable shift with value 2
    // Your code here
}

Hint

Use shift := 2 to create the variable.

3

Apply bitwise operations

Create variables andResult, orResult, xorResult, notResult, leftShiftResult, and rightShiftResult to store the results of a & b, a | b, a ^ b, ^a, a << shift, and a >> shift respectively.

DSA Go

package main

import "fmt"

func main() {
    a := 12
    b := 5
    shift := 2
    // Calculate bitwise AND, OR, XOR, NOT, left shift, and right shift
    // Your code here
}

Hint

Use the operators &, |, ^, ^ (NOT), <<, and >> with the variables.

4

Print the results

Print the values of andResult, orResult, xorResult, notResult, leftShiftResult, and rightShiftResult each on a new line using fmt.Println.

DSA Go

package main

import "fmt"

func main() {
    a := 12
    b := 5
    shift := 2

    andResult := a & b
    orResult := a | b
    xorResult := a ^ b
    notResult := ^a
    leftShiftResult := a > shift

    // Print all results
    // Your code here
}

Hint

Use fmt.Println to print each variable on its own line.

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