Question

我想写一个treeFold函数，它接受：'a - ＆gt;类型的函数f b - ＆gt; a，类型a的值x，名为t的树B，返回类型a的值。通过执行树的有序遍历来计算返回值，通过x传递部分结果。这是我的代码：

import Control.Exception
import Control.Monad
import Control.DeepSeq
import qualified Data.List as List
import Test.HUnit

data Tree a  =  Empty
             |  Node a (Tree a) (Tree a)
             deriving (Show, Eq)



insertTree :: ( Ord a, Show a ) => Tree a -> a -> Tree a
insertTree Empty x  =  Node x Empty Empty
insertTree ( Node v tLeft tRight ) x
    | x == v = Node v tLeft tRight
    | x < v = Node v (insertTree tLeft x) tRight
    | x > v = Node v tLeft (insertTree tRight x)


createTree :: ( Ord a, Show a ) => [ a ] -> Tree a
createTree = foldl insertTree Empty

intTree = createTree [9,7,2,8,6,0,5,3,1]

listTree = createTree ( List.permutations [ 0 .. 3 ] )

strTree = createTree [ "hello"
                     , "world"
                     , "lorem"
                     , "ipsum"
                     , "dolor"
                     , "sit"
                     , "amet"
                     ]
treeFold :: (a -> b -> b -> b) -> b -> Tree a -> b
treeFold f z Empty = z
treeFold f z (Node v l r) = f v (subfold l) (subfold r)
    where subfold = foldTree f z

但是当我运行代码时，我得到了“无法匹配的类型错误”。我想知道如何解决这个问题？例如：主要＆gt; treeFold（+）10 intTree，而不是获得Main＆gt; 51，我得到的不能匹配类型错误。非常感谢任何帮助。

Answer 1

在你的功能中

treeFold :: (a -> b -> b -> b) -> b -> Tree a -> b
treeFold f z Empty = z
treeFold f z (Node v l r) = f v (subfold l) (subfold r)
    where subfold = foldTree f z

您对三个值使用f：v，(subfold l)和(subfold r)。这就是您的类型签名需要f来获取三个参数的原因。

在我看来，你最好两次使用两个参数f，一次合并v和(subfold l)，另一个合并{{1} }}：

(subfold r)

这确实意味着，如果我们假设treeFold f z Empty = z treeFold f z (Node v l r) = f (f v (subfold l)) (subfold r) where subfold = treeFold f z，那么f :: a -> b -> c因为subfold l :: b，而且f v (subfold l)因为它是从subfold l :: c返回的。因此，treeFold和c属于同一类型（b）和c ~ b。这被用作第一个f v (subfold l) :: b的第一个参数，所以也f。因此对于这个新的use-f-two版本，

a ~ b

运行我的foldTree函数时出错

1 个答案: