Check if Number is Even or Odd Using Bits in DSA C++ - Build from Scratch

Check if Number is Even or Odd Using Bits

📖 Scenario: Imagine you are building a simple program that helps a cashier quickly check if a number is even or odd using a fast method. This can be useful in many places where quick decisions are needed based on numbers.

🎯 Goal: You will create a program that uses bit operations to check if a number is even or odd and then prints the result.

📋 What You'll Learn

Create an integer variable named number with the value 29

Create an integer variable named bit_mask with the value 1

Use a bitwise AND operation between number and bit_mask to check if the number is even or odd

Print "Odd" if the number is odd, otherwise print "Even"

💡 Why This Matters

🌍 Real World

Bitwise operations are used in low-level programming, embedded systems, and performance-critical applications to quickly check or manipulate data.

💼 Career

Understanding bitwise operations helps in roles like embedded software development, systems programming, and optimizing algorithms.

1

Create the number variable

Create an integer variable called number and set it to 29.

DSA C++

# Your code here

Hint

Use int number = 29; to create the variable.

2

Create the bit mask variable

Create an integer variable called bit_mask and set it to 1.

DSA C++

int number = 29;
# Your code here

Hint

Use int bit_mask = 1; to create the mask for checking the last bit.

3

Check if the number is even or odd using bitwise AND

Use an if statement to check if number & bit_mask is equal to 1. If yes, the number is odd; otherwise, it is even.

DSA C++

int number = 29;
int bit_mask = 1;
// Your code here

Hint

Use if ((number & bit_mask) == 1) to check the last bit.

4

Print the result

Inside the if block, print "Odd". Inside the else block, print "Even".

DSA C++

int number = 29;
int bit_mask = 1;
if ((number & bit_mask) == 1) {
    // Print "Odd" here
} else {
    // Print "Even" here
}

Hint

Use std::cout << "Odd" << std::endl; and std::cout << "Even" << std::endl; to print the result.

Practice

1. Which bit operation can be used to check if a number is even or odd?

easy

A. Check the least significant bit using AND with 1

B. Shift the number right by 1

C. XOR the number with 1

D. Invert all bits of the number

Solution

1. Step 1: Understand the least significant bit (LSB)

   The LSB of a binary number determines if it is even or odd. If LSB is 0, number is even; if 1, number is odd.

2. Step 2: Use AND operation with 1 to check LSB

   Performing bitwise AND with 1 isolates the LSB. Result 0 means even, 1 means odd.

3. Final Answer:

   Check the least significant bit using AND with 1 -> Option A

4. Quick Check:

   LSB & 1 = 0 (even) or 1 (odd) [OK]

Hint: Use n & 1 to check even or odd quickly [OK]

Common Mistakes:

• Using XOR instead of AND to check LSB
• Checking the most significant bit instead
• Using bit shift without checking bits

2. Which of the following C++ code snippets correctly checks if num is odd using bits?

easy

A. if ((num & 1) == 0) { /* odd */ }

B. if ((num >> 1) == 1) { /* odd */ }

C. if ((num ^ 1) == 1) { /* odd */ }

D. if ((num & 1) == 1) { /* odd */ }

Solution

1. Step 1: Understand bitwise AND with 1

   Using (num & 1) extracts the least significant bit. If it equals 1, number is odd.

2. Step 2: Check condition for odd number

   The condition (num & 1) == 1 correctly identifies odd numbers.

3. Final Answer:

   if ((num & 1) == 1) { /* odd */ } -> Option D

4. Quick Check:

   (num & 1) == 1 means odd [OK]

Hint: Use (num & 1) == 1 to check odd numbers [OK]

Common Mistakes:

• Using == 0 to check odd (actually checks even)
• Using bit shift without comparison
• Using XOR incorrectly for odd check

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

int num = 14;
if (num & 1) {
    std::cout << "Odd";
} else {
    std::cout << "Even";
}

medium

A. Odd

B. Even

C. 14

D. Error

Solution

1. Step 1: Evaluate bitwise AND of num and 1

   14 in binary is 1110. The least significant bit is 0. So, 14 & 1 = 0.

2. Step 2: Check the if condition

   Since (num & 1) is 0, the else block executes, printing "Even".

3. Final Answer:

   Even -> Option B

4. Quick Check:

   14 & 1 = 0 means Even [OK]

Hint: Check LSB: 0 means even, 1 means odd [OK]

Common Mistakes:

• Assuming 14 is odd because it ends with 4
• Confusing bitwise AND with arithmetic AND
• Printing num instead of "Even" or "Odd"

4. Identify the error in the following code snippet that checks if a number is odd:

int num = 7;
if (num & 1 == 1) {
    std::cout << "Odd";
} else {
    std::cout << "Even";
}

medium

A. Incorrect variable type for num

B. Using == instead of =

C. Missing parentheses around (num & 1)

D. No error, code is correct

Solution

1. Step 1: Analyze operator precedence in condition

   In C++, == has higher precedence than &. So, num & 1 == 1 is parsed as num & (1 == 1), which is num & 1. This is not the intended comparison.

2. Step 2: Identify the error

   Without parentheses, the expression does not explicitly compare (num & 1) to 1, which can cause confusion or unintended behavior. Parentheses are needed: (num & 1) == 1.

3. Final Answer:

   Missing parentheses around (num & 1) -> Option C

4. Quick Check:

   Operator precedence requires parentheses for clarity [OK]

Hint: Use parentheses: (num & 1) == 1 to avoid precedence issues [OK]

Common Mistakes:

• Misinterpreting operator precedence (== > &)
• Thinking parentheses are not needed
• Believing the code works correctly without parentheses

5. You want to write a function in C++ that returns true if a number is even and false if odd, using bitwise operations. Which of the following implementations is correct?

hard

A. bool isEven(int n) { return (n & 1) == 0; }

B. bool isEven(int n) { return (n & 1) == 1; }

C. bool isEven(int n) { return (n >> 1) == 0; }

D. bool isEven(int n) { return !(n ^ 1); }

Solution

1. Step 1: Understand even number bit pattern

   Even numbers have least significant bit 0. So (n & 1) == 0 means even.

2. Step 2: Check each function implementation

   bool isEven(int n) { return (n & 1) == 1; } returns true for odd numbers, so incorrect. bool isEven(int n) { return (n & 1) == 0; } correctly returns true for even numbers. Options C and D do not correctly check evenness.

3. Final Answer:

   bool isEven(int n) { return (n & 1) == 0; } -> Option A

4. Quick Check:

   Even if LSB is 0 [OK]

Hint: Return (n & 1) == 0 for even check [OK]

Common Mistakes:

• Returning true for odd numbers by mistake
• Using bit shift incorrectly for even check
• Using XOR without understanding its effect