Question

我之前已经定义了一个函数，该函数获取Maybe的列表并将其转换为列表的floop :: [Maybe a] -> Maybe [a] floop [] = Just [] floop (Nothing:_) = Nothing floop (Just x:xs) = fmap (x:) $ floop xs，如下所示：

foldr

现在我想重新定义它以与更大类的容器兼容，而不仅仅是列表，我发现它需要实现函数mappend，mempty，{{1 }}，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)

As（我认为）它确保为给定容器实现这些功能，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)

在调查之后，我发现Monoid的种类与Functor和Foldable的种类不同，但我不明白为什么会出现这种情况，也不知道如何纠正错误。

对于那些感兴趣的人，这是当前的实施：

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：我已经意识到这已经作为内置函数（sequence）存在，但我打算将其作为一个学习练习来实现。

Answer 1

Monoidal applicatives由Alternative类描述，使用(<|>)和empty代替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

Answer 2

这可能是提出hoogle的好地方。

搜索t (m a)-> m (t a)返回sequenceA :: (Traversable t, Applicative f) => t (f a) -> f (t a)作为第一个结果。然后，这将导致Traversable类型类与您正在寻找的类型非常接近。

正如Lee所说，你可以使用Alternative类，它类似于Monoid。稍微宽泛一点：

sequence' :: (Alternative t, Foldable t, Applicative a) => t (a b) -> a (t b)
sequence' = foldr step (pure empty)
  where step = liftA2 prepend
        prepend = (<|>) . pure

这里prepend首先将一些单个元素包装到t中，这样它就可以使用（＆lt; |＆gt;）来预先添加它。 liftA2让我们抽象出应用程序a，您可以将它想象为解开两个参数，将它们应用于前置并将结果包装回来。

将类型参数限制为Monoid

2 个答案: