Well, its more direct in that, while it increases the number of branches, it minimizes the number of statements executed on any branch. It's also the solution that maps most directly to the problem statement, and, absent a strong technical reason to do otherwise, a direct mapping from requirements to code is a good thing.
I disagree that it maps most directly to the problem statement. You're performing a common factor computation in your mind, which may be more difficult given numbers other than 3 and 5. In my opinion, pattern matching offers the most direct solution and comes with an abundance of compiler optimizations. Here's an example in Rust...
for i in range(1i, 101) {
match (i % 3, i % 5) {
(0, 0) => println!("Fizzbuzz"),
(0, _) => println!("Fizz"),
(_, 0) => println!("Buzz"),
_ => println!("{}", i),
}
}
> I disagree that it maps most directly to the problem statement. You're performing a common factor computation in your mind, which may be more difficult given numbers other than 3 and 5.
Well, sure, explicitly calling out i % 15 rather than (i % 3) && (i % 5) or the equivalent has that problem.
> In my opinion, pattern matching offers the most direct solution and comes with an abundance of compiler optimizations.
Pattern matching is not available in many languages, but, sure, where its available, its a great choice. Note that this still has a distinct case for the case where both % 3 and % 5 are true, rather than just testing those cases independently and sequentially and concatenating, so I think it falls into the general class of solutions I was describing.
The solution in Haskell is quite clean, I believe.
fizzBuzz n
| n `mod` 15 == 0 = "FizzBuzz"
| n `mod` 3 == 0 = "Fizz"
| n `mod` 5 == 0 = "Buzz"
| otherwise = show n
main = mapM_ (print . fizzBuzz) [1..100]
I agree with you about generalizing pattern matching for less simple cases. Your example brought to mind view patterns, about which Oliver O'Charles had a nice writeup recently [1]. Nifty little extension.
let buzzer number =
match number with
| i when i % 3 = 0 && i % 5 = 0 -> "FizzBuzz"
| i when i % 3 = 0 -> "Fizz"
| i when i % 5 = 0 -> "Buzz"
| i -> (sprintf "%i" i)
for i = 1 to 100 do
printfn "%s" (buzzer i)
