Count even and odd numbers with while loop in C++

The provided C++ code is designed to count the number of even and odd numbers entered by the user, excluding the terminating 0.

Code

/*
* Program to count even and odd numbers.
 *
 * This program prompts the user to enter a sequence of integers, ending the sequence with a 0.
 * It then counts the number of even and odd numbers entered (excluding the final 0) and displays the counts.
 *
 * How it works:
 * 1. The program initializes two counters for even and odd numbers.
 * 2. It prompts the user to enter a number and reads the user input.
 * 3. If the number is not 0, it checks if the number is even or odd and increments the respective counter.
 * 4. The program repeats steps 2 and 3 until the user enters 0.
 * 5. Finally, it displays the counts of even and odd numbers.
 *
 * Note: The program considers 0 as neither even nor odd for the purpose of this count.
 */

#include <iostream>
using namespace std;

int main() {
    int evenCount = 0; // Counter for even numbers
    int oddCount = 0;  // Counter for odd numbers
    int userInput;     // Variable to store the user's input

    cout << "Enter a number: ";
    cin >> userInput;

    while (userInput != 0) {
        if (userInput % 2 == 1)
            oddCount++; // Increment odd counter if the number is odd
        else
            evenCount++; // Increment even counter if the number is even

        cout << "Enter a number: ";
        cin >> userInput;
    }

    cout << "Even numbers : " << evenCount << endl; // Display the count of even numbers
    cout << "Odd numbers : " << oddCount << endl;   // Display the count of odd numbers

    return 0;
}

Explanation

The provided C++ code is designed to count the number of even and odd numbers entered by the user, excluding the terminating 0. The program operates in a loop, prompting the user to input integers until a 0 is entered, which signals the end of input. It utilizes the modulo operator (%) to distinguish between even and odd numbers.

Initially, the program declares and initializes two integer variables, evenCount and oddCount, to zero. These variables serve as counters for the even and odd numbers, respectively.

int evenCount = 0; // Counter for even numbers
int oddCount = 0;  // Counter for odd numbers

The program then enters a loop, first prompting the user to enter a number. This is achieved using cout for the prompt and cin to read the user’s input into the variable userInput.

cout << "Enter a number: ";
cin >> userInput;

Within the loop, the program checks if the input is not 0. If it’s not, it determines whether the number is even or odd by using the modulo operation (userInput % 2). If the result is 1, the number is odd, and oddCount is incremented. Otherwise, the number is even, and evenCount is incremented.

if (userInput % 2 == 1)
    oddCount++; // Increment odd counter if the number is odd
else
    evenCount++; // Increment even counter if the number is even

This process repeats until the user inputs 0, at which point the loop terminates. Finally, the program outputs the counts of even and odd numbers using cout.

cout << "Even numbers : " << evenCount << endl;
cout << "Odd numbers : " << oddCount << endl;

This code snippet effectively demonstrates basic C++ input/output operations, conditional statements, and loop control structures, making it a straightforward example for developers familiar with C++ but new to this specific logic.

Output

Enter a number: 13
Enter a number: 212
Enter a number: 345
Enter a number: 23
Enter a number: 0
Even numbers : 1
Odd numbers : 3

Process finished with exit code 0