Odd word problem: Difference between revisions

Line 1,121:

While it seems like this solution would break the task's rules, Haskell is non-strict, therefore this yields the same behavior of reading and printing one character at a time, without excess storage into a "string". To prove it, run the program and manually enter the input string (Windows command prompt does not respect buffering settings, but urxvt on on Linux does).

<lang Haskell>import System.IO

(BufferMode(..), getContents, hSetBuffering, stdin, stdout)

import Data.Char (isAlpha)

isAlpha :: Char -> Bool

isAlpha = flip elem $ ['a'..'z'] ++ ['A'..'Z']

split :: String -> (String, String)

split = ~~break $ not .~~ isAlpha

split = span isAlpha

parse :: String -> String

parse [] = []

parse l =

let (a, w) = split l

(b, x) = splitAt 1 w

(c, y) = split x

(d, z) = splitAt 1 y

in a ++ b ++ reverse c ++ d ++ parse z

in a <> b <> reverse c <> d <> parse z

main :: IO ()

main =

~~main~~ = hSetBuffering stdin NoBuffering >> hSetBuffering stdout NoBuffering >>

hSetBuffering stdin NoBuffering >> hSetBuffering stdout NoBuffering >> getContents >>=

~~getContents >>=~~ putStr . (takeWhile (/= '.')) . parse >~~> putStrLn "."</lang~~>

putStr . takeWhile (/= '.') . parse >>

putStrLn "."</lang>

If the above is not acceptable, or if Haskell was implicitly strict, then this solution would satisfy the requirements:

<lang Haskell>isAlpha :: Char -> Bool