FizzBuzz Program in C++

This program prints numbers from 1 to 100, replacing multiples of 3 with “Fizz”, multiples of 5 with “Buzz”, and multiples of both with “FizzBuzz”.

Program Explanation

The program follows these steps:

1. Loop through numbers from 1 to 100 using a for loop.
2. For each number, check if it is a multiple of both 3 and 5 by using the modulo operator %. If true, print “FizzBuzz”.
3. If the number is not a multiple of both, check if it is a multiple of 3. If true, print “Fizz”.
4. If the number is not a multiple of 3, check if it is a multiple of 5. If true, print “Buzz”.
5. If the number is neither a multiple of 3 nor 5, print the number itself.

C++ Program Code

// FizzBuzz program in C++

#include <iostream>

int main() {
    // Loop through numbers from 1 to 100
    for (int i = 1; i <= 100; ++i) {
        // Check if the number is a multiple of both 3 and 5
        if (i % 3 == 0 && i % 5 == 0) {
            std::cout << "FizzBuzz" << std::endl;
        }
        // Check if the number is a multiple of 3
        else if (i % 3 == 0) {
            std::cout << "Fizz" << std::endl;
        }
        // Check if the number is a multiple of 5
        else if (i % 5 == 0) {
            std::cout << "Buzz" << std::endl;
        }
        // If none of the above, print the number itself
        else {
            std::cout << i << std::endl;
        }
    }
    return 0;
}
    

Explanation of the C++ Code

• #include <iostream>: This is the preprocessor directive that includes the input-output stream library.
• int main(): This is the main function where the program execution begins.
• for (int i = 1; i <= 100; ++i): This is the for loop that iterates from 1 to 100.
• if (i % 3 == 0 && i % 5 == 0): This condition checks if the current number is a multiple of both 3 and 5.
• std::cout << "FizzBuzz" << std::endl;: If the number is a multiple of both 3 and 5, print “FizzBuzz”.
• else if (i % 3 == 0): This condition checks if the current number is a multiple of 3.
• std::cout << "Fizz" << std::endl;: If the number is a multiple of 3, print “Fizz”.
• else if (i % 5 == 0): This condition checks if the current number is a multiple of 5.
• std::cout << "Buzz" << std::endl;: If the number is a multiple of 5, print “Buzz”.
• else: This handles the case where the number is neither a multiple of 3 nor 5.
• std::cout << i << std::endl;: Print the number itself.
• return 0;: Return 0 indicates that the program ended successfully.