Ruby FizzBuzz using the ternary operator two times

Question

Task: Implement a function which loops from 1 to 100 and prints "Fizz" if the counter is divisible by 3. Prints "Buzz" if the counter is divisible by 5 and "FizzBuzz", when the counter is divisible by 3 and 5.

The instructor gave me the hint, that it's solvable using the ternary operator. Otherwise I would have used an if-then-else approach.

My solution:

def fizz_buzz
  for i in 1...100
    puts i % 3 == 0 ? i % 5 == 0 ? "FizzBuzz" : "Fizz" : i % 5 == 0 ? "Buzz" : i
  end
end

What's your opinion about the approach? Would you still consider it as readable?

I personally consider the readability hard on the border.

Which approach would you prefer and why?

Answer 1

Ternary statement is cleaner and often faster to comprehend than if/else; especially true for languages with ceremony syntax, such as curly braces. And I like nested ternary statements (or is it singular?) but it's important to format to help the reader with logic flow. Assume the reader knows the operator structure and resist over-indentation resulting eye boggling pseudo if/else.

def fizz_buzz
  for i in 1...100
    puts i % 3 == 0 ?  i % 5 == 0 ? "FizzBuzz" : "Fizz"   : 
         i % 5 == 0 ? "Buzz" : i
  end
end

Answer 2

No nested if's (and no ternary operators):

def fizzbuzz
  (1..100).each do |n|
    res = ""
    res << "Fizz" if n % 5 == 0
    res << "Buzz" if n % 3 == 0
    res << n.to_s if res.empty?
    puts res
  end
end

With one extra line of code this transforms to a FizzBuzzBazz.

Answer 3

In all the cases except the one where performance is a top priority, I'd prefer the most readable approach:

def fizz_buzz
  100.times do |i|
    if i % 3 == 0 && i % 5 == 0
      puts "FizzBuzz"
    elsif i % 3 == 0
      puts "Fizz"
    elsif i % 5 == 0
      puts "Buzz"
    else
      puts i
    end
  end
end

This code immediately tells you that there are 4 cases, you may easily see the conditions for each of them, all the outcomes are clear and the code is still very simple. Other programmers (or future you) will spend almost no time to read and understand it.

Some people will say that this code is not optimal because it does more checks than the required minimum, but that's usually much less important than readability. It'll certainly take several additional nanoseconds, but in most cases it is negligible. For example, in web development (where Ruby is mostly used), typical response times are in tens or hundreds of milliseconds. Developer time, however, costs a lot. Caring about code readability pays

Also, notice that i % 3 == 0 && i % 5 == 0 is the same as i % 15 == 0, but I'd suggest not to make this replacement either, because i % 3 == 0 && i % 5 == 0 is more intent-revealing

I suggest reading "99 bottles of OOP" by Sandi Metz for a deeper perspective on that.

Answer 4

A post just to suggest a different formatting of radarbob's suggestion

def fizz_buzz
  for i in 1...100
    puts i % 3 == 0 ? i % 5 == 0 ? "FizzBuzz" : "Fizz"
                    : i % 5 == 0 ? "Buzz" : i
  end
end