问题 强制模式顺序

主要问题是你希望有一个排列而不是单个值 view。我们只有一种允许排列的类型 - 记录。

所以，我们可以添加新数据，记录类型

data B = F|T -- just shorter name for Bool in patterns
data Palette = P {isW, isU, isB, isR, isG :: B}

bool2b :: Bool -> B
bool2b True  = T
bool2b False = F

viewColors :: Card -> Palette
viewColors (Card colors) = let m = bool2b . (`member` colors)
    in P {isW = m W, isU = m U, isB = m B, isR = m R, isG = m G}

foo :: Card -> String
foo (viewColors -> P {isW=T, isB=T}) = "card is white and black"
foo _ = "whatever"

更新

我们也可以 拒绝 错误的模式。但是这个解决方案更难看，但它允许使用“经典”模式

{-# LANGUAGE GADTs #-}
{-# LANGUAGE EmptyDataDecls #-}
{-# LANGUAGE RankNTypes #-}
data Color = W | U | B | R | G  deriving (Eq)

data W' 
data U' 
data B'
data R'
data G'

data Color' a where
      W' :: Color' W'
      U' :: Color' U'
      B' :: Color' B'
      R' :: Color' R'
      G' :: Color' G'

data M a = N | J a -- just shorter name for Maybe a in patterns

data Palette = Palette 
      (M (Color' W')) 
      (M (Color' U')) 
      (M (Color' B')) 
      (M (Color' R')) 
      (M (Color' G'))

并定义 viewColor：

viewColors :: Card -> Palette
viewColors (Card colors) = 
  let 
    m :: Color -> Color' a -> M (Color' a)
    m c e = if c `member` colors then J e else N
  in P (m W W') (m U U') (m B B') (m R R') (m G G')

foo :: Card -> String
foo (viewColors -> Palette (J W') N (J B') N N) = 
      "card is white and black"
foo _ = "whatever"

编辑：进一步的测试表明，该解决方案实际上并不起作用。

你实际上不需要任何更多的扩展，我想出了一个可以做你想要的解决方案，但你可能想要优化它，重命名一些东西，并使它变得不那么难看。您只需要创建一个新的数据类型并实现 Eq 你自己并让操作员使用 infixr：

{-# LANGUAGE ViewPatterns #-}
import Data.Set

data Color = W | U | B | R | G
    deriving (Show, Eq, Ord)

data Card = Card (Set Color) -- simplified Card type with only its colors

-- you may need to fiddle with the precedence here
infixr 0 :*
data MyList a = END | a :* (MyList a) deriving (Show)

myFromList :: [a] -> MyList a
myFromList [] = END
myFromList (x:xs) = x :* myFromList xs

instance Eq a => Eq (MyList a) where
    END == END = True
    END == _   = False
    _   == END = False
    l1  == l2  = allElem l1 l2 && allElem l2 l1
        where
            -- optimize this, otherwise it'll just be really slow
            -- I was just too lazy to write it correctly
            elemMyList :: Eq a => a -> MyList a -> Bool
            elemMyList a ml = case ml of
                END -> False
                (h :* rest) -> if a == h then True else elemMyList a rest
            allElem :: Eq a => MyList a -> MyList a -> Bool
            allElem END l = True
            allElem (h :* rest) l = h `elemMyList` l && allElem rest l

viewColors :: Card -> MyList Color
viewColors (Card colors) = myFromList $ toList colors

fuz :: Card -> String
fuz (viewColors -> (W :* END)) = "it's white"
fuz (viewColors -> (W :* U :* END)) = "it's white and blue"
fuz (viewColors -> (W :* B :* END)) = "it's white and black"
fuz (viewColors -> (W :* B :* R :* END)) = "it's white, black, and red"
fuz (viewColors -> (W :* U :* B :* R :* G :* END)) = "it's all colors"
fuz _ = "I don't know all my colors"

main = do
    putStrLn $ fuz $ Card $ fromList [W, B]
    putStrLn $ fuz $ Card $ fromList [B, W]

编辑：刚刚修改了一下代码

我认为你应该专注于准确地表达卡片的颜色，然后担心其他问题，比如稍后简化。这听起来像你的 Bool 元组解决方案几乎是完美的，但我猜这张卡片 必须 有一种颜色，对吗？

在这种情况下，这样的东西可能会起作用，并且很容易进行模式匹配：

data CardColors = W' BlackBool GreenBool ...
                | B' WhiteBool GreenBool ...
                | G' BlackBool WhiteBool ...
                ....

data BlackBool = B 
               | NotB
-- etc.

您可以非常轻松地创建具有已定义顺序的异构列表，但我不认为这种多态将在此为您服务。

（不是你问题的答案，但希望能解决你的问题！）

我会选择可能有效的最蠢的事情：

is :: Card -> Color -> Bool
is card col = col `elem` (viewColors card) -- can be optimized to use the proper elem!

接着

foo :: Card -> String
foo c
    | c `is` B && c `is` W = "card is black and white"
    | c `is` R || c `is` G = "card is red or green"
    | otherwise = "whatever"

如果拼出整个列表来检查一张卡片是否有全部5种颜色太长，那么你可以定义额外的组合器

hasColors :: Card -> [Color] -> Bool
hasColors card = all (`elem` (viewColors card))

这是不可接受的吗？