Question

我有一个树的以下定义

 data Tree a = Leaf a
           | Node [Tree a] 
     deriving (Show)

以下可折叠的实例：

 instance Foldable (Tree) where
 foldMap f (Leaf t) = (f t)
 foldMap f (Node t) = (foldMap `mappend` (foldMap f) t)

这段代码引发了我的错误

 Couldn't match type `a' with `Tree a'
  `a' is a rigid type variable bound by
      the type signature for
        foldMap :: Monoid m => (a -> m) -> Tree a -> m
      at trees.hs:8:5
 Expected type: [a]
   Actual type: [Tree a]

如何在Tree a类型的实例声明中创建t而不是？

Answer 1

（我假设您确实在实例声明中缩进了代码，否则编译器会对此抱怨。）

问题在于：

 foldMap f (Node t) = (foldMap `mappend` (foldMap f) t)

什么是（foldMap`mappend` ...）应该做的？您将foldMap本身视为monadic值。我想你想做的只是foldMap (foldMap f) t。

顺便说一句，GHC可以自动为您导出可折叠实例（以及Functor和Traversable）。只需写下

{-# LANGUAGE DeriveFunctor, DeriveFoldable #-}
data Tree a = Leaf a
            | Node [Tree a]
  deriving (Show, Functor, Foldable)

可折叠和树木

1 个答案: