如何在Haskell中实现数据类型的自定义排序？

Question

Let's say I have a data type that represents a deck of poker cards like so

data Suit = Clubs | Spades | Hearts | Diamonds deriving (Eq)

instance Show Suit where
    show Diamonds = "♦"
    show Hearts = "♥"
    show Spades = "♠"
    show Clubs = "♣"

data Value = Two | Three | Four | Five | Six | Seven | Eight | Nine | Ten | Jack | Queen | King | Ace deriving (Eq, Ord)

data Card = Card {
      value :: Value
    , suit :: Suit
    } deriving (Eq, Ord)

instance Show Card where show (Card vs) = show s ++ show v

Haskell helps me already a lot by allow me to derive Eq and Ord without explicitly specifying those relations. This is especially useful, because I want to use Ord to ultimately compare the value of two hands according to Poker rules.

Now in Poker, the suits do not really matter in terms of ordering. Therefore I tried

instance Ord Suit where
    compare Clubs Spades = EQ
    compare Clubs Hearts = EQ
    compare Clubs Diamonds = EQ
    compare Spades Hearts = EQ
    compare Spades Diamonds = EQ
    compare Hearts Diamonds = EQ

This is already a bit verbose ... and it does not even work:

*P054> a
♠A
*P054> b
♣A
*P054> a < b
*** Exception: P054.hs:(12,9)-(17,36): Non-exhaustive patterns in function compare

So how can I properly define an ordering on Suit that expresses the fact that all suits are equal?

Answer 1

You are missing several combinations, eg all with Clubs on the right hand side. If all are equal, what are the possible results of compare , regardless of the input? There is only one: EQ . Therefore we don't even need to look at a or b :

instance Ord Suit where
    compare _ _ = EQ

However, that Ord instance is rather useless. Alternatively you can create a custom instance for Card ,

instance Ord Card where
    compare a b = compare (value a) (value b)
    -- compare = compare `on` value -- using `on` from Data.Function
    -- compare = comparing value    -- using `comparing` from Data.Ord