2

我正在尝试编写函数，使用 monoids 和 Foldable 将列表中的所有元素相加和相乘。我设置了一些我认为有效的代码：

data Rose a = a :> [Rose a]
    deriving (Eq, Show)

instance Functor Rose where
    fmap f rose@(a:>b) = (f a :> map (fmap f) b) 

class Monoid a where
    mempty ::           a
    (<>)   :: a -> a -> a

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

newtype Sum     a = Sum     { unSum     :: a } deriving (Eq, Show)
newtype Product a = Product { unProduct :: a } deriving (Eq, Show)

instance Num a => Monoid (Sum a) where
    mempty           = Sum 0
    Sum n1 <> Sum n2 = Sum (n1 + n2)

instance Num a => Monoid (Product a) where
    mempty                   = Product 1
    Product n1 <> Product n2 = Product (n1 * n2)

class Functor f => Foldable f where
    fold    :: Monoid m =>             f m -> m
    foldMap :: Monoid m => (a -> m) -> f a -> m
    foldMap f a = fold (fmap f a)

instance Foldable [] where
    fold = foldr (<>) mempty

instance Foldable Rose where
    fold (a:>[]) = a <> mempty
    fold (a:>b)  = a <> (fold (map fold b))

然后在定义了不同的 Foldable 实例以及 Sum 和 Product 类型之后，我想定义两个函数，它们分别相加将数据结构中的元素相乘，但这会产生我不知道如何解释的错误，我必须承认我与实际逻辑相比，猜测工作更多，因此欢迎对您的答案进行全面解释。

fsum, fproduct :: (Foldable f, Num a) => f a -> a
fsum b     = foldMap Sum b
fproduct b = foldMap Product b

错误：

Assignment3.hs:68:14: error:
    * Occurs check: cannot construct the infinite type: a ~ Sum a
    * In the expression: foldMap Sum b
      In an equation for `fsum': fsum b = foldMap Sum b
    * Relevant bindings include
        b :: f a (bound at Assignment3.hs:68:6)
        fsum :: f a -> a (bound at Assignment3.hs:68:1)
   |
68 | fsum b     = foldMap Sum b
   |              ^^^^^^^^^^^^^

Assignment3.hs:69:14: error:
    * Occurs check: cannot construct the infinite type: a ~ Product a
    * In the expression: foldMap Product b
      In an equation for `fproduct': fproduct b = foldMap Product b
    * Relevant bindings include
        b :: f a (bound at Assignment3.hs:69:10)
        fproduct :: f a -> a (bound at Assignment3.hs:69:1)
   |
69 | fproduct b = foldMap Product b
   |              ^^^^^^^^^^^^^^^^^
4

1 回答 1

4

如果在 中使用Sum（或Product）foldMap，您将首先将中的项目映射Foldable到Sums（或Products）。因此fsumwill - 就像你定义的那样 - 的结果是 a Sum a，而不是 a a：

fsum :: (Foldable f, Num a) => f a -> Sum a
fsum b = foldMap Sum b

为了检索包装在Sum构造函数中的值，您可以使用unSum :: Sum a -> agetter 获取它：

fsum :: (Foldable f, Num a) => f a -> a
fsum b = unSum (foldMap Sum b)

或在eta-reduction之后：

fsum :: (Foldable f, Num a) => f a -> a
fsum = unSum . foldMap Sum

对于 a 也应该发生同样的情况Product。

于 2018-10-12T12:13:45.080 回答