用于泛型Haskell树的通用foldr和foldl？

问题描述

$ b <$ p

$ p> 的第三个参数。 code>

foldr 期望其第三个参数的类型为 [c]

所以你需要处理 ts 到类型为 [c] ，并将结果传递给 foldr 而不是 ts 本身。 功能应该是一个好地方开始。

How can I write generalized foldr and foldl function for generic Haskell trees, given this definition?

data (Eq a, Show a) => Tree a = Void | Node a [Tree a]
    deriving (Eq, Show)

treefoldr :: (Eq a, Show a) => 
   (a -> b -> c) -> c -> (c -> b -> b) -> b -> Tree a -> c

treefoldl :: (Eq a, Show a) =>
   (b -> a -> c) -> c -> (c -> b -> b) -> b -> Tree a -> c

Even if I can understand how foldr and foldl functions work in Haskell, I'm not quite sure how to write this generalized function for trees.

EDIT: I tried something like this (not even compiling):

treefoldr  _ g1 _ _    Void       = g1
treefoldr f1 g1 f2 g2 (Node a ts) = f1 a (foldr f2 g2 ts)

EDIT 2: another try...

treefoldr _ z1 _ _   Void      = z1
treefoldr f z1 g z2 (Node a ts) =
   f a (foldr g z2 (map (\x -> treefoldr f z1 g z2 x) ts))

treefoldl _ z1 _ _   Void      = z1
treefoldl f z1 g z2 (Node a ts) =
   f (foldl g z2 (map (\x -> treefoldl f z1 g z2 x) ts)) a

treefoldr is working, however treefoldl not:

解决方案

The error message in full:

Couldn't match expected type `c' against inferred type `Tree a'
  `c' is a rigid type variable bound by
      the type signature for `treefoldr' at so.hs:5:14
  Expected type: [c]
  Inferred type: [Tree a]
In the third argument of `foldr', namely `ts'
In the second argument of `f1', namely `(foldr f2 g2 ts)'

That means that

• ts is of type [Tree a]
• you are using it as the third argument to foldr
• foldr expects its third argument to be of type [c]
• [c] and [Tree a] are different types, hence this is an error

So you need to process ts into something of type [c] and pass that result to foldr instead of ts itself. The map function would be a good place to start.

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