Question

我尝试为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个函数调用彼此递归，但我得到一些编译错误。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：

名为.hs：27：34：无法匹配预期类型b0 -> c0' with actual type Tree Int 可能的原因：Node' is applied to too many arguments In the first argument of（。）＆＃39;，即`Node n fst＆＃39; 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

有什么问题，我应该如何解决这个问题？

Answer 1

这一行

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))

尝试组合树而不是函数，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，请注意上面的代码两次计算numberForest (n+1) xs，导致指数爆炸。你可以避免这种情况，使用let ... in ...如下

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

您可以使用let：

中的模式匹配进一步改进此功能

numberTree n (Node _ (x:xs)) =  let (forest, n') = numberForest (n+1) xs
                                in (Node n forest, n')

Answer 2

状态monad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

这比明确的状态传递更具可读性。

试图迭代森林Haskell编译错误

2 个答案: