无法将类型同义词与 Either 匹配

Question

我是初学者Haskell。但不太确定 Either 在模式匹配中是如何工作的。

这是我的代码：

type Rank = Either Pip Court
type Pip = Int
type Deck = [Card]
data Card = Joker | Card Suit Rank
data Court = Ace | Jack | Queen | King deriving (Show, Eq, Ord)
data Suit = Hearts | Diamonds | Clubs | Spades deriving (Show, Ord, Eq)

snap :: Card -> Card -> String
snap Joker Joker = "SNAP"
snap (Card s1 r1) (Card s2 r2)
    | r1 == r2 = "SNAP"
snap _ _ = "..."

GHCi 告诉我 Couldn't match type ‘Court’ with ‘Either Pip Court’，有人可以帮我吗？

Answer 1

Either 不仅仅是类型的联合；它是一个 tagged 联合，这意味着每个值都必须明确指定包装值出现在哪个 "side" 类型上。这是一个示例，（使用为您的 Card 类型派生的 Show 实例）：

*Main> Card Hearts Jack

<interactive>:3:13: error:
    • Couldn't match type ‘Court’ with ‘Either Pip Court’
      Expected type: Rank
        Actual type: Court
    • In the second argument of ‘Card’, namely ‘Jack’
      In the expression: Card Hearts Jack
      In an equation for ‘it’: it = Card Hearts Jack

Card 需要一个类型为 Either Pip Court 的参数，但你给了它一个普通的 Court 值。使用显式换行：

*Main> Card Hearts (Right Jack)
Card Hearts (Right Jack)

这同样适用于创建编号卡片：

*Main> Card Hearts 3

<interactive>:5:13: error:
    • No instance for (Num Rank) arising from the literal ‘3’
    • In the second argument of ‘Card’, namely ‘3’
      In the expression: Card Hearts 3
      In an equation for ‘it’: it = Card Hearts 3
*Main> Card Hearts (Left 3)
Card Hearts (Left 3)

您不需要更改 snap 的定义，因为 Either 已经有一个 Eq 实例，只要两个包装类型都具有 Eq实例本身；它认为任何 Right 值不等于任何 Left 值，并且如果包装值相等，则匹配的 Right 或 Left 值相等。

*Main> snap (Card Hearts (Right Jack)) (Card Hearts (Left 3))
"..."
*Main> snap (Card Hearts (Right Jack)) (Card Spades (Right Jack))
"SNAP"

无法将类型同义词与 Either 匹配

Couldn't match type Synonym with Either

haskell

types

pattern-matching

either