[All/Scala] Trier de coordonnées dans un tableau à deux dimensions

[niahoo]

Autre implémentation naïve en Haskell

-- Module Skouares

data Square = Square { x :: Int

                     , y :: Int

                     , user :: String

                     } deriving (Show)



square :: Int -> Int -> String -> Square

square x y user = Square {x=x,y=y,user=user}



squareX :: Int -> [Square]

squareX cx = filter ((cx==).x) squares



squareXY :: Int -> Int -> [Square]

squareXY cx cy = filter (((cx,cy)==).xy) squares



xy :: Square -> (Int, Int)

xy a = (x a, y a)



squares :: [Square]

squares = [ square 2 2 "oxman"

          , square 4 6 "oxman"

          , square 4 7 "Sephi-Chan"

          , square 2 7 "Sephi-Chan"

          , square 4 4 "oxman"

          , square 2 7 "Sephi-Chan"

          ]

La même chose sans les records, du coup on crée la fonction x comme accessor en plus

-- Module Skouares

data Square = Square Int Int String deriving (Show)



squareX :: Int -> [Square]

squareX cx = filter ((cx==).x) squares



squareXY :: Int -> Int -> [Square]

squareXY cx cy = filter (((cx,cy)==).xy) squares



xy :: Square -> (Int, Int)

xy (Square x y u) = (x, y)



x :: Square -> Int

x (Square x y u) = x



squares :: [Square]

squares = [ Square 2 2 "oxman"

          , Square 4 6 "oxman"

          , Square 4 7 "Sephi-Chan"

          , Square 2 7 "Sephi-Chan"

          , Square 4 4 "oxman"

          , Square 2 7 "Sephi-Chan"

          ]