數據類型MyList的實例聲明為Class Semigroup和Class Monoid

Question

MyList就像列表一樣，我已經定義了一些函數來創建concat' （相當於(++) 。 我想為MyList a做一個實例聲明

instance Semigroup [a] where
    -- (<>) :: [a] -> [a] -> [a]
    (<>) = (++)
instance Monoid [a] where
    -- mempty :: [a]
    mempty = []

我有GHCi版本8.2.2。

data MyList a = Leer | Element a (MyList a) deriving (Show,Eq)

last' :: Eq a => MyList a -> a
last' Leer = undefined
last' (Element x(xs))
    | xs == Leer = x
    | otherwise = last' xs
w :: MyList a -> a -> MyList a
w Leer x = Element x (Leer)
w xs x = Element x (xs)
concat' :: Eq a => MyList a -> MyList a -> MyList a
concat' x Leer = x
concat' (Element x (xs)) ys
    | xs == Leer = (Element x (ys))
    | otherwise = concat' (Element x (deletelst xs)) (w ys (last' xs))

length' :: MyList a -> Int
length' Leer = 0
length' (Element x (xs)) = 1 + length' (xs)
take' :: Int -> MyList a -> MyList a
take' _ Leer = Leer
take' 0 _ = Leer
take' z (Element x (xs)) = Element x (take' (z-1) xs)

deletelst :: MyList a-> MyList a
deletelst Leer = Leer
deletelst xs = (take' ((length' xs) - 1) xs) 

instance Semigroup (MyList a) where
    (<>) = concat'
instance Monoid (MyList a) where
    mempty = Leer
    mappend = (<>)

當我嘗試編譯程序時，會彈出此錯誤消息：

FH.hs:115:12: error:
    * No instance for (Eq a) arising from a use of concat'
    Possible fix: add (Eq a) to the context of the instance declaration
    * In the expression: concat'
    In an equation for `<>': (<>) = concat'
    In the instance declaration for `Semigroup (MyList a)'

Answer 1

錯誤消息告訴您問題出在哪里 ：

* In the expression: concat'
In an equation for `<>': (<>) = concat'
In the instance declaration for `Semigroup (MyList a)'

在這里：

instance Semigroup (MyList a) where
    (<>) = concat'

它還告訴您問題是什么 ：

* No instance for (Eq a) arising from a use of concat'
Possible fix: add (Eq a) to the context of the instance declaration

確實，您正在使用concat'來定義(<>) ，但是您對concat'的定義需要一個Eq a實例，該實例不作為Semigroup (MyList a)實例聲明的上下文Semigroup (MyList a) -換句話說，您定義instance Semigroup (MyList a)而不是instance Eq a => Semigroup (MyList a) 。 您可以添加該上下文，但是有一個更原則的解決方案：通過使用模式匹配，從concat' （以及last' ）的定義中刪除（不必要的） Eq a約束：

concat' :: MyList a -> MyList a -> MyList a
concat' x Leer              = x
concat' (Element x Leer) ys = Element x ys
concat' (Element x xs) ys   = concat' (Element x (deletelst xs)) (w ys (last' xs))

last' :: MyList a -> a
last' Leer             = undefined
last' (Element x Leer) = x
last' (Element x xs)   = last' xs