haskell - 将类型参数限制为 Monoid

Question

我之前定义了一个函数，它接受一个列表 Maybe s 并将其变成 Maybe一个列表，像这样:

floop :: [Maybe a] -> Maybe [a]
floop [] = Just []
floop (Nothing:_) = Nothing
floop (Just x:xs) = fmap (x:) $ floop xs

现在我想重新定义它以兼容更大类的容器，而不仅仅是列表，我发现它需要实现功能foldr , mappend , mempty , fmap , 和 pure ;所以我认为以下类型行是合适的:

floop :: (Foldable t, Functor t, Monoid t) => t (Maybe a) -> Maybe (t a)

因为(我认为)它确保为给定的容器实现了这些功能，但是它会导致以下错误:

Expecting one more argument to ‘t’
The first argument of ‘Monoid’ should have kind ‘*’,
  but ‘t’ has kind ‘* -> *’
In the type signature for ‘floop'’:
  floop' :: (Foldable t, Functor t, Monoid t) =>
            t (Maybe a) -> Maybe (t a)

查了一下，发现Monoid的种类与Functor的种类不同和 Foldable ，但我不明白为什么会这样，也不明白如何纠正错误。

对于那些感兴趣的人，这是当前的实现:

floop :: (Foldable t, Functor t, Monoid t) => t (Maybe a) -> Maybe (t a)
floop xs = let
                f :: (Foldable t, Functor t, Monoid t) => Maybe a -> Maybe (t a) -> Maybe (t a)
                f Nothing _ = Nothing
                f (Just x) ys = fmap (mappend $ pure x) ys
            in
                foldr f (Just mempty) xs

注意:我已经知道这已经作为一个内置函数存在 ( sequence )，但我打算将它作为一个学习练习来实现。

Answer 1

幺半群应用由 Alternative 描述类，使用 (<|>)和 empty而不是 mappend和 mempty :

floop :: (Foldable t, Alternative t) => t (Maybe a) -> Maybe (t a)
floop xs = let
                f :: (Foldable t, Alternative t) => Maybe a -> Maybe (t a) -> Maybe (t a)
                f Nothing _ = Nothing
                f (Just x) ys = fmap ((<|>) $ pure x) ys
            in
                foldr f (Just empty) xs