[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)