FizzBuzz in C++

FizzBuzz is the most famous beginner programming exercise — and a question interviewers really do ask to check you can write a simple loop. The task sounds trivial, but it quietly tests loops, conditionals, and the modulo operator all at once. Let’s build it from scratch.

The Rules

Print every number from 1 to 100, with three twists:

• If a number is a multiple of 3, print Fizz instead.
• If it’s a multiple of 5, print Buzz instead.
• If it’s a multiple of both 3 and 5, print FizzBuzz.
• Otherwise, just print the number.

That’s it. The whole challenge is turning those four rules into code in the right order.

Step 1: Loop From 1 to 100

Before worrying about Fizz or Buzz, get a loop printing the plain numbers:

#include <iostream>

int main() {
    for (int i = 1; i <= 100; ++i) {
        std::cout << i << "\n";
    }
    return 0;
}

This prints 1 through 100, one per line. Now we just need to replace some of those numbers with words.

Step 2: Test Divisibility With Modulo

The modulo operator % gives the remainder after division. If the remainder is zero, the number divides evenly. So i % 3 == 0 means “i is a multiple of 3.” That single test is the heart of FizzBuzz.

The trick is the order of your checks. A multiple of both 3 and 5 is a multiple of 15, so you must test for 15 first. If you don’t, i % 3 == 0 will match first and you’ll print Fizz when you wanted FizzBuzz.

The Complete FizzBuzz Program

Putting the loop and the ordered checks together:

#include <iostream>

int main() {
    for (int i = 1; i <= 100; ++i) {
        if (i % 15 == 0)
            std::cout << "FizzBuzz\n";
        else if (i % 3 == 0)
            std::cout << "Fizz\n";
        else if (i % 5 == 0)
            std::cout << "Buzz\n";
        else
            std::cout << i << "\n";
    }
    return 0;
}

Because these are else if branches, only the first matching one runs — which is exactly why checking 15 first works. Run it and you’ll see 1 2 Fizz 4 Buzz Fizz 7 ... all the way to 100.

A Cleaner Version Without Checking 15

There’s an elegant alternative that avoids the “test 15 first” rule entirely. Build up a string: add Fizz for a multiple of 3, add Buzz for a multiple of 5, and if nothing was added, use the number itself.

#include <iostream>
#include <string>

int main() {
    for (int i = 1; i <= 100; ++i) {
        std::string output;
        if (i % 3 == 0) output += "Fizz";
        if (i % 5 == 0) output += "Buzz";
        if (output.empty()) output = std::to_string(i);
        std::cout << output << "\n";
    }
    return 0;
}

For a multiple of 15, both if checks fire, so output becomes "FizzBuzz" automatically. This version is easy to extend — add a Bazz rule for multiples of 7 and you don’t have to rethink any combined cases.

Common Mistakes

The number-one bug is checking 3 or 5 before 15 — you’ll never see FizzBuzz. The second is using = (assignment) instead of == (comparison) inside the if. And remember % only works on integers; it’s perfect here since every value is a whole number.

Related Articles

• C++ Modulo Operator — the % operator that powers FizzBuzz
• C++ Conditionals Tutorial — if, else if, and else explained
• C++ Loops Tutorial — for and while loops in depth
• C++ Even or Odd Program — another modulo classic
• C++ Ternary Operator — tidy up small conditionals

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