Question

我是学习Haskell的初学者。但不确定Either在模式匹配中如何工作。

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不仅仅是类型的并集；这是一个 tagged 联合，这意味着每个值都必须明确指定发生包装值的类型的哪一侧。这是一个示例，（具有为您的Show类型派生的Card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Either Pip Court类型的参数，但是您给了它一个简单的Court值。使用显式包装：

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

创建数字卡也是如此：

*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Eq已经有一个Either实例，只要两个包装类型本身都具有Eq实例;它认为任何Right值都不等于任何Left值，并且如果包装的值相等，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"

无法将同义词与任何一个匹配

1 个答案: