[icedog]

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),  
            }  
        }

[dragonwriter]

> 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.

[icedog]

I'll take an extra branch over concatenation and dealing with new line chars any day.

[qubitcoder]

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.

[1] https://ocharles.org.uk/blog/posts/2014-12-02-view-patterns....

[ghuntley]

Using F# pattern matching:

    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)

[tupshin]

If only rust had (optional only, please) fall through on matches, then you could skip a whole line :D