FizzBuzz

Posted by okkez Sat, 26 May 2007 02:03:00 GMT
Haskell
import System main = putStr $ unlines [fb(x) | x <- [1..100]] fb n | n `mod` 15 == 0 = "FizzBuzz" | n `mod` 5 == 0 = "Buzz" | n `mod` 3 == 0 = "Fizz" | otherwise = show n
Erlang
-module(fizzbuzz). -export([fizzbuzz/0,fb/1]). fizzbuzz() -> [ fb(X) || X <- lists:seq(1,100)]. fb(N) when N rem 15 =:= 0 -> "FizzBuzz"; fb(N) when N rem 5 =:= 0 -> "Buzz"; fb(N) when N rem 3 =:= 0 -> "Fizz"; fb(N) -> N.

Erlang の方の動かし方。(インストールは略)

  • erl シェルを起動
1> c(fizzbuzz.erl). %=> コンパイル {ok,fizzbuzz} 2> fizzbuzz:fizzbuzz. %=> 関数呼び出しの括弧は必要 ** 1: illegal expression ** 3> fizzbuzz:fizzbuzz(). [1, 2, "Fizz", 4, "Buzz", "Fizz", 7, 8, "Fizz", "Buzz", 11, "Fizz", 13, 14, "FizzBuzz", 16, 17, "Fizz", 19, "Buzz", "Fizz", 22, 23, "Fizz", "Buzz", 26, "Fizz", 28, 29|...] %=> リストの出力は省略される 4>

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