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Discussion on: Unconditional Challenge: FizzBuzz without `if`

yujiri8 profile image
Ryan Westlund

Well, it still uses && which you said wasn't allowed...

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avaq profile image
Aldwin Vlasblom

You are referring to the following clause:

Do this without "secret" conditionals like || and && (in loosely typed languages like JavaScript)

My understanding is that this applies to the usage of such operators as conditionals when they are applied to non-boolean types. For example:

const fizz = n % 3 === 0 && 'Fizz' || String(n)

In my code, it is used strictly on Boolean values, which I don't think was against the rules.

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kallmanation profile image
Nathan Kallman Author

It uses it in a strictly boolean sense, which I'll allow. What isn't allowed is the loose/early-return way JS can use &&; a contrived example that would not pass hard mode:

const fizzbuzz = n => (n % 3 && (n % 5 && n || "Buzz")) || (n % 5 && "Fizz" || "FizzBuzz")

(an easy litmus test: does changing the order of the && expression change the result?)