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我尝试为森林的每个节点分配一个数字，这样就不会有 2 个节点具有相同的数字。我尝试使用 2 个相互调用递归的函数，但我得到了一些编译错误。这是代码：

numberTree :: Int -> Tree a -> (Tree Int,Int)

numberTree n (Node c []) = (Node n [], n+1)
numberTree n (Node _ (x:xs)) =  (Node n  fst.numberForest  (n+1) xs, snd.numberForest   (n+1) xs)


numberForest :: Int -> [Tree a] -> ([Tree Int],Int)

numberForest n (x:xs) = ((fst(numberTree n x )):(fst(numberForest (n+1) xs)), snd(numberForest (n+1) xs)+1)

numberForest n (x:xs) = ((fst(numberTree n x )):(fst(numberForest (n+1) xs)), snd(numberForest (n+1) xs)+1)

我得到的错误是：

.hs:27:34: 无法匹配预期类型b0 -> c0' with actual typeTree Int 可能原因：( Node' is applied to too many arguments In the first argument of.)'，即 `Node n fst' 在表达式中：Node n fst 。numberForest (n + 1) xs

.hs:27:34:
Couldn't match expected type `Tree Int' with actual type `a0 -> c0
In the expression: Node n fst . numberForest (n + 1) xs
In the expression:
  (Node n fst . numberForest (n + 1) xs,
   snd . numberForest (n + 1) xs)
In an equation for `numberTree':
    numberTree n (Node _ (x : xs))
      = (Node n fst . numberForest (n + 1) xs,
         snd . numberForest (n + 1) xs)

 .hs:27:42:
Couldn't match type `(a1, b1) -> a1' with `[Tree Int]'
Expected type: Forest Int
  Actual type: (a1, b1) -> a1
Probable cause: `fst' is applied to too few arguments
In the second argument of `Node', namely `fst'
In the first argument of `(.)', namely `Node n fst'

 .hs:27:46:
Couldn't match expected type `a0 -> b0'
            with actual type `([Tree Int], Int)'
Possible cause: `numberForest' is applied to too many arguments
In the second argument of `(.)', namely `numberForest (n + 1) xs'
In the expression: Node n fst . numberForest (n + 1) xs

.hs:27:70:
Couldn't match expected type `Int' with actual type `a2 -> c1'
In the expression: snd . numberForest (n + 1) xs
In the expression:
  (Node n fst . numberForest (n + 1) xs,
   snd . numberForest (n + 1) xs)
In an equation for `numberTree':
    numberTree n (Node _ (x : xs))
      = (Node n fst . numberForest (n + 1) xs,
         snd . numberForest (n + 1) xs)

.hs:27:74:
Couldn't match expected type `a2 -> (a3, c1)'
            with actual type `([Tree Int], Int)'
Possible cause: `numberForest' is applied to too many arguments
In the second argument of `(.)', namely `numberForest (n + 1) xs'
In the expression: snd . numberForest (n + 1) xs

出了什么问题，我应该如何解决这个问题？

4

2 回答 2

4

这条线

numberTree n (Node _ (x:xs)) =  (Node n  fst.numberForest  (n+1) xs, snd.numberForest   (n+1) xs)

实际上意味着

numberTree n (Node _ (x:xs)) =  ( (Node n fst) . (numberForest (n+1) xs)
                                , snd . (numberForest (n+1) xs))

它试图组合树而不是函数，导致编译器抱怨。你可能想要这样的东西：

numberTree n (Node _ (x:xs)) =  ( Node n  (fst (numberForest  (n+1) xs))
                                , snd (numberForest (n+1) xs))

但是，请注意上面的代码计算numberForest (n+1) xs了两次，导致指数级爆炸。您可以避免这种情况，例如使用let ... in ...如下

numberTree n (Node _ (x:xs)) =  let result = numberForest (n+1) xs
                                in (Node n (fst result), snd result)

您可以在以下内容中使用模式匹配进一步改进let：

numberTree n (Node _ (x:xs)) =  let (forest, n') = numberForest (n+1) xs
                                in (Node n forest, n')
于 2014-12-11T14:44:11.203 回答
0

状态单子是你的朋友。

import Control.Monad.Trans.State

data Tree   a = Node a (Forest a)
type Forest a = [Tree a]

numberTreeS :: Tree a -> State Int (Tree Int)
numberTreeS (Node _ xs) = do
    n <- get
    put (n + 1)
    xs' <- numberForestS xs
    return $ Node n xs'

numberForestS :: Forest a -> State Int (Forest Int)
numberForestS = mapM numberTreeS

numberForest :: Forest a -> Forest Int
numberForest xs = evalState (numberForestS xs) 0

这比显式状态传递更具可读性。

于 2014-12-11T16:52:17.027 回答