C++ Program To Check Whether Number Is Even Or Odd

This article explores how to determine if a given number is even or odd using C++. You will learn the fundamental logic behind this check and practical C++ implementations.

Problem Statement

A number is considered even if it is perfectly divisible by 2, leaving no remainder. Conversely, a number is odd if it is not perfectly divisible by 2, meaning it leaves a remainder of 1 when divided by 2. Accurately identifying even or odd numbers is a basic but crucial operation in various programming scenarios.

Example

Let's quickly see how an even/odd check works:

Input: 7 Output: 7 is an Odd number.

Input: 10 Output: 10 is an Even number.

Background & Knowledge Prerequisites

To understand the solutions presented, you should have a basic grasp of:

• C++ syntax: How to write simple C++ programs.
• Input/Output operations: Using cout for output and cin for input.
• Conditional statements: if-else structures.
• Arithmetic operators: Specifically, the modulo operator (%).
• Bitwise operators: The bitwise AND operator (&) for an alternative approach.

Use Cases or Case Studies

Checking if a number is even or odd has numerous practical applications:

• Array and List Processing: Alternate styling for elements (e.g., zebra striping in tables), processing elements at even/odd indices.
• Game Development: Determining turns (Player 1 on odd turns, Player 2 on even turns), movement patterns, or specific game logic based on numerical parity.
• Data Validation: Ensuring specific input adheres to an even or odd requirement.
• Mathematical Algorithms: Many number theory problems and algorithms rely on parity checks.
• User Interface (UI) Design: Highlighting rows in a list differently for better readability (e.g., even rows have a light grey background).
• Resource Allocation: Distributing tasks evenly or handling specific resources based on whether a task ID is even or odd.

Solution Approaches

We will explore two common and effective methods to check for even or odd numbers in C++.

Approach 1: Using the Modulo Operator (%)

The most straightforward way to check for even or odd is by using the modulo operator, which returns the remainder of a division.

Summary: A number n is even if n % 2 equals 0; otherwise, it's odd.

// Check Even or Odd using Modulo Operator
#include <iostream>

int main() {
    int number;

    // Step 1: Prompt the user to enter a number
    std::cout << "Enter an integer: ";
    std::cin >> number;

    // Step 2: Use the modulo operator to check for even or odd
    // If the remainder when divided by 2 is 0, it's even.
    // Otherwise, it's odd.
    if (number % 2 == 0) {
        std::cout << number << " is an Even number." << std::endl;
    } else {
        std::cout << number << " is an Odd number." << std::endl;
    }

    return 0;
}

Run

Sample Output:

Enter an integer: 42
42 is an Even number.

Enter an integer: 17
17 is an Odd number.

Stepwise Explanation:

1. Include iostream: This line brings in the necessary library for input and output operations.
2. Declare Variable: An integer variable number is declared to store the user's input.
3. Get Input: The program prompts the user to enter an integer and reads it into the number variable using std::cin.
4. Apply Modulo Operator: The core logic number % 2 == 0 is evaluated.
   • number % 2 calculates the remainder when number is divided by 2.

1. Conditional Output:
   • If the condition (number % 2 == 0) is true, the if block executes, printing that the number is even.

Approach 2: Using the Bitwise AND Operator (&)

A more performance-optimized approach, especially for embedded systems or low-level programming, involves using the bitwise AND operator. This method checks the Least Significant Bit (LSB) of the number.

Summary: In binary representation, an even number always has its LSB as 0, while an odd number always has its LSB as 1. Performing number & 1 isolates the LSB.

// Check Even or Odd using Bitwise AND Operator
#include <iostream>

int main() {
    int number;

    // Step 1: Prompt the user to enter a number
    std::cout << "Enter an integer: ";
    std::cin >> number;

    // Step 2: Use the bitwise AND operator to check the LSB
    // If (number & 1) is 0, the LSB is 0, meaning it's even.
    // If (number & 1) is 1, the LSB is 1, meaning it's odd.
    if ((number & 1) == 0) {
        std::cout << number << " is an Even number." << std::endl;
    } else {
        std::cout << number << " is an Odd number." << std::endl;
    }

    return 0;
}

Run

Sample Output:

Enter an integer: 8
8 is an Even number.

Enter an integer: 13
13 is an Odd number.

Stepwise Explanation:

1. Include iostream: Standard input/output library.
2. Declare Variable: An integer variable number is declared for input.
3. Get Input: The program takes an integer input from the user.
4. Apply Bitwise AND: The key operation is (number & 1).
   • Any integer N can be represented in binary. 1 in binary is ...0001.

1. Conditional Output:
   • If (number & 1) == 0 is true, the number is even.

Conclusion

Determining whether a number is even or odd is a foundational concept in programming. Both the modulo operator (%) and the bitwise AND operator (&) provide simple and effective solutions in C++. While the modulo operator is generally more intuitive and readable for beginners, the bitwise AND approach offers a slightly more efficient alternative for specific use cases.

Summary

• An even number is perfectly divisible by 2 (remainder 0).
• An odd number is not perfectly divisible by 2 (remainder 1).
• Modulo Operator (%): number % 2 == 0 for even, number % 2 != 0 for odd. This is the most common and readable method.
• Bitwise AND Operator (&): (number & 1) == 0 for even, (number & 1) != 0 (or == 1) for odd. This method checks the Least Significant Bit (LSB) and can be slightly more performant.
• Both methods provide reliable ways to classify numbers by parity.