警卫

Question

哈斯克尔替代类和虚函数可以说我有以下ADT，

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

我已经做了以下情况吧，

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) 

到目前为止好。让我们看看我们可以迄今为止做，

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 

给人，

*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和预期的单子绑定都在工作。现在我想使用alternative中的guard函数。基本上我想根据一些谓词修剪pColors和pNationality。所以我尝试定义，

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

但是，这给了我一个错误，我认为（请纠正我，如果我错了），这是因为Haskell是解释[]从a不同。但我认为a可能是任何东西，而[]是其中之一，所以问题是什么？

Answer 1

但我认为a可以是任何东西和[]是任何一个，所以有什么问题？

不是说a可以是任何东西，但是a必须能够是任何东西。这是如果用户写入x :: Property [a]; x = empty，那必须被允许，但是x :: Property Int; x = empty也必须被允许。你的定义只允许前者，所以这是不正确的。

没有办法定义有效empty您Property类型，除非你让它产生错误（如empty = Property "" undefined）或改变的Property定义添加一个案例的空属性。