Haskell類的誤解

Question

你好，我需要創建一個類，其中包含一個方法prize'計算一些分數。我想實例的情況下，我得到一個Order，以後我將不得不再比如如果我得到一個[Order]，這是代碼：

class Ped a where 
    prize' :: Fractional b => a -> b 

instance (Integral n, Fractional p) => Ped (Order n p) where 
    prize' x = prizeOrder x 

data (Integral c,Fractional p) => Product c p 
    = Prod c String p 
    deriving (Show, Read) 

data (Integral n, Fractional p) => Order n p 
    = PdMult (Product n p) n 
    | PdUnit (Product n p) 
    deriving (Show, Eq) 

prize :: (Fractional p, Integral c) =>(Product c p) -> p 
prize (Prod _ _ x) = x 

prizeOrder :: (Fractional p, Integral c) => (Order c p) -> p 
prizeOrder (PdMult p n) = (prize p) * (fromIntegral n) 
prizeOrder (PdUnit p) = prize p 

前奏說：

Could not deduce (p ~ b) 
from the context (Integral n, Fractional p) 
bound by the instance declaration 
at src\Funciones.hs:6:10-55 
or from (Fractional b) 
bound by the type signature for 
    prize' :: Fractional b => Order n p -> b 
at src\Funciones.hs:7:5-11 
`p' is a rigid type variable bound by 
    the instance declaration 
    at src\Funciones.hs:6:10 
`b' is a rigid type variable bound by 
    the type signature for prize' :: Fractional b => Order n p -> b 
    at src\Funciones.hs:7:5 
Expected type: Order b 
    Actual type: Order n p 
Relevant bindings include 
    x :: Order n p 
    (bound at src\Funciones.hs:7:13) 
    prize' :: Order n p -> b 
    (bound at src\Funciones.hs:7:5) 
In the first argument of `prizeOrder', namely `x' 
In the expression: prizeOrder x 

Answer 1

的問題是您的class定義給出了額外的自由度，您的instance未能提供。在你的類，你聲明：

class Ped a where 
    prize' :: Fractional b => a -> b 

這意味着程序員可以選擇他/她想要什麼Fraction b爲b。

現在如果我們查看到您的實例：

instance (Integral n, Fractional p) => Ped (Order n p) where 
    prize' x = prizeOrder x 

在這裏，您說prize'取決於prizeOrder。然而，prizeOrder函數簽名：

prizeOrder :: (Fractional p, Integral c) => (Order c p) -> p 

這意味着你不能選擇b可言。如果我使用prize'，並且預期返回類型爲Float，那麼Ped將會接受此結果，但不會由數據結構Order Double Int接受。

使用多參數類可以「解決」這個，你把b在class簽名：

class Ped a b where 
    prize' :: a -> b 

接下來，你需要用第二類參數定義實例，以保證輸出爲p：

instance (Integral n, Fractional p) => Ped (Order n p) p where 
    prize' x = prizeOrder x 

您需要激活一些額外的GHC特點，充分鱈魚Ë會讀到這樣：

{-# LANGUAGE MultiParamTypeClasses #-} 
{-# LANGUAGE FlexibleInstances #-} 
{-# LANGUAGE DatatypeContexts #-} 

class Ped a b where 
    prize' :: a -> b 

instance (Integral n, Fractional p) => Ped (Order n p) p where 
    prize' x = prizeOrder x 

data (Integral c,Fractional p) => Product c p 
    = Prod c String p 
    deriving (Show, Read) 

data (Integral n, Fractional p) => Order n p 
    = PdMult (Product n p) n 
    | PdUnit (Product n p) 
    deriving (Show) 

prize :: (Fractional p, Integral c) =>(Product c p) -> p 
prize (Prod _ _ x) = x 

prizeOrder :: (Fractional p, Integral c) => (Order c p) -> p 
prizeOrder (PdMult p n) = (prize p) * (fromIntegral n) 
prizeOrder (PdUnit p) = prize p 

但是，作爲@bheklilr說，不要使用「類約束的數據類型」。此外，最好問他/她是否支付使用class。