FizzBuzz in C#

The comments on this Hacker News post generated a bunch of opinions on the ability for developers to solve a FizzBuzz problem. Specifically in a coding interview. I’ve never actually seen what the FizzBuzz problem was, so I had to look it up. In this blog post I solve FizzBuzz in C#.

The FizzBuzz problem is a challenge for a developer to write a method, in any language, to return different parts of the word FizzBuzz.

It goes like this:

• If a number is divisible by 3 or 5 => return FizzBuzz
• If a number is only divisible by 3 => return Fizz
• If a number is divisible by 5 => return Buzz
• If a number is not divisible by 3 or 5 => return the number

Over twenty years of programming experience and I’ve never come across question in an interview so I thought it would be fun to jump onto .Net Fiddle to give it a try. In this function you take any number and check if it can be divisible by 3 or 5. The trick to this function is to use the % mod operator to check if a number is divisible by another number. The mod operator will return a number if there is a remainder after division between the two supplied numbers. If the number returned after division using the mod operator you will return zero – meaning there isn’t a remainder after division. This is what the function is all about. Here’s my FizzBuzz implementation I wrote this morning after reading the Hacker News post.

Simple FizzBuzz implementation in C#

public class Program
{
	public static void Main()
	{
		int i = 0;
		while(i <=100)
		{
			string fizzWord = GetFizz(i);
			Console.WriteLine(fizzWord);
			i = i + 1;
		}
		
	}
	
	public static string GetFizz(int i)
	{
		if(i % 3 == 0 && i % 5==0)
		{
		  return "FizzBuzz";
		}
		else if(i % 3 == 0)
		{
			return "Fizz";
		}
		else if(i % 5 == 0)
		{
			return "Buzz";
		}
		
		return i.ToString();
	}
}