警衛

Question

哈斯克爾替代類和虛函數可以說我有以下ADT，

data Properties a = Property String a 
    deriving (Show,Eq) 

我已經做了以下情況吧，

instance Functor Properties where 
    fmap f (Property name prop) = Property name (f prop) 
-- 
instance Applicative (Properties) where 
    pure a = Property "" a 
    (Property _ f) <*> p = fmap f p 

instance Monad (Properties) where 
    return a = Property "" a 
    (Property name prop) >>= f = (f prop) 

到目前爲止好。讓我們看看我們可以迄今爲止做，

pColors = Property "Color" ["Blue", "Red", "White"] 
pNationality = Property "Nationality" ["Italian", "Norwegian", "Spanish"] 

answers = do 
    color <- fmap permutations pColors 
    nation <- fmap permutations pNationality 
    return $ zip color nation 

給人，

*Main> answers 
Property "" [(["Blue","Red","White"],["Italian","Norwegian","Spanish"]),(["Red","Blue","White"],["Norwegian","Italian","Spanish"]),(["White","Red","Blue"],["Spanish","Norwegian","Italian"]),(["Red","White","Blue"],["Norwegian","Spanish","Italian"]),(["White","Blue","Red"],["Spanish","Italian","Norwegian"]),(["Blue","White","Red"],["Italian","Spanish","Norwegian"])] 

所以fmap和預期的單子綁定都在工作。現在我想使用alternative中的guard函數。基本上我想根據一些謂詞修剪pColors和pNationality。所以我嘗試定義，

instance Alternative (Properties) where 
    empty = Property "" [] 

但是，這給了我一個錯誤，我認爲（請糾正我，如果我錯了），這是因爲Haskell是解釋[]從a不同。但我認爲a可能是任何東西，而[]是其中之一，所以問題是什麼？

Answer 1

但我認爲a可以是任何東西和[]是任何一個，所以有什麼問題？

不是說a可以是任何東西，但是a必須能夠是任何東西。這是如果用戶寫入x :: Property [a]; x = empty，那必須被允許，但是x :: Property Int; x = empty也必須被允許。你的定義只允許前者，所以這是不正確的。

沒有辦法定義有效empty您Property類型，除非你讓它產生錯誤（如empty = Property "" undefined）或改變的Property定義添加一個案例的空屬性。