Haskell重構建議

Question

我正在尋找一些關於以下代碼的重構/最佳實踐建議。我想盡量避免擴展，同時爲不同的「版本」保持獨立的模塊，這些模塊是互斥的。我目前的解決方案是使用一個類並使用存在性量化爲每個CountrySets創建一個通用類型。

如果我使用面向對象的方式，這對我來說很容易完成，但是我無法想到「功能」。

謝謝你的時間。

Province.hs

{-# LANGUAGE ExistentialQuantification, RankNTypes #-} 
module Province where 

    class Country a where 
    --some stuff that is not important 

    data Power = 
    forall a. (Show a, Eq a, Country a) => Power a | 
    Netural | 
    Water 

    data Unit = Unit { 
    power :: forall a. (Show a, Eq a, Country a) => a, 
    piece :: Piece 

    data Piece = Tank | Plane 

    data Province = Province { 
    power :: Power, 
    provName :: String 
    } deriving (Eq) 

    instance Show Power where 
    show (Power b) = "Power " ++ show b 
    show (Netural) = "Netural" 
    show (Water) = "Water" 

    instance Eq Power where 
    (==) a b = Prelude.show a == Prelude.show b 

Version1.hs

import Province 

    data CountrySet1 = 
    Country11 | 
     Country12 
    deriving (Eq, Show) 
    instance Country CountrySet1 where 

    provs = 
    one1:one2:[] 

    one1 = Province (Power Country11) "Place11" 
    one2 = Province (Power Country12) "Place12" 

Version2.hs

import Province 

    data CountrySet2 = 
    Country21 | 
    Country22 
     deriving (Eq, Show) 
    instance Country CountrySet2 where 

    provs = 
    two1:two2:[] 

    two1 = Province (Power Country11) "Place21" 
    two2 = Province (Power Country12) "Place22" 

Answer 1

你不必把類構造函數在數據類型中您可以參數化，而不是在a可變數據類型，這樣就可以將制約該類型的類實例自己，就像這樣：

-- Note that I added a type variable to "Power" 
data Power a = Power a | Neutral | Water 

instance (Show a) => Show (Power a) where ... 

instance (Eq a) => Eq (Power a) where ... 

...或者你可以做大多數人一樣，用deriving ：

data Power a = Power a | Neutral | Water deriving (Eq, Show) 

這會產生你寫的完全一樣的情況下（除了Eq一個會比你寫的更有效）。不需要擴展！

然後，如果你想a是一個特定的類型，你只是這麼說！

-- Version1.hs 
myValue1 :: Power CountrySet1 
myValue1 = ... 


-- Version2.hs 
myValue2 :: Power CountrySet2 
myValue2 = ... 

這些都是完全兼容的，兩個實現可以並存在一起。