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我正在尝试附加适用于嵌套列表的函数，如常规列表。我想使用 Either String (Nested a) 以便它返回错误或附加列表。但它一直在失败。我没有在任何地方做 NestedList[NestedList a]。为什么它说它期望 [NestedList (NestedList a)]

module Main where

data NestedList a=Elem a | List[NestedList a] deriving Show

flatten ::NestedList a->[a]
flatten (Elem x)=[x]
flatten (List(x:xs))=flatten(x)++flatten(List xs)
--flatten NestedList (x:xs)=flatten(x)++flatten(xs)
flatten(List [])=[]

------------------

count::[a]->Int
count a=length (a)

-----------------
append::NestedList a->NestedList a->Either String (NestedList a)
append (_) (Elem a)=Left "Elements are not allowed"
append (Elem a) (_)=Left "Elements are not allowed"
append (List a) (List b)=Right (List (flatten(List a)++flatten(List b)))
-------------------

main=do
        print(flatten (List[Elem 1, Elem 2,List[Elem 1, Elem 2 ]]))
        print(count(flatten (List[Elem 1, Elem 2,List[Elem 1, Elem 2 ]])))
        print(append (List[List[Elem 1, Elem 2 ]]) (List[Elem 1, Elem 2,List[Elem 1, Elem 2 ]]    ))

给我这个错误

   flatten_list.hs:18:52:
   Couldn't match type `a' with `NestedList a'
      `a' is a rigid type variable bound by
      the type signature for
        append :: NestedList a
                  -> NestedList a -> Either String (NestedList a)
      at flatten_list.hs:15:9
       Expected type: [NestedList (NestedList a)]
  Actual type: [NestedList a]
In the first argument of `List', namely `a'
In the first argument of `flatten', namely `(List a)'
In the first argument of `(++)', namely `flatten (List a)'

flatten_list.hs:18:69:
    Couldn't match type `a' with `NestedList a'
  `a' is a rigid type variable bound by
      the type signature for
        append :: NestedList a
                  -> NestedList a -> Either String (NestedList a)
      at flatten_list.hs:15:9
    Expected type: [NestedList (NestedList a)]
  Actual type: [NestedList a]
In the first argument of `List', namely `b'
In the first argument of `flatten', namely `(List b)'
In the second argument of `(++)', namely `flatten (List b)'
4

1 回答 1

1 
import Data.Traversable

append :: NestedList a -> NestedList a -> Either String (NestedList a)
append (Elem x) (Elem y) = Right $ List [Elem x, Elem y] 
append (Elem _) _ = Left ""
append _ (Elem _) = Left "" 
append (List xs) (List ys) = fmap List $ sequenceA $ zipWith append xs ys

您在问题中编写的附加似乎只想展平并连接两个列表，这是毫无意义的，因为您正在丢弃所有结构。如果这是所需的行为，则只需使用常规列表：

append' :: NestedList a -> NestedList a -> Either String [a]
append' (Elem x) (Elem y) = Right $ [x,y]
append' (Elem _) _ = Left ""
append' _ (Elem _) = Left "" 
append' a b        = Right $ flatten a ++ flatten b

您还可以根据free monad定义您的数据类型：

import Control.Monad.Free
type NestedList = Free []

flatten :: NestedList a -> [a]
flatten = retract 

append :: NestedList a -> NestedList a -> Either String (NestedList a)
append (Pure x) (Pure y) = Right $ Free [Pure x, Pure y] 
append (Pure _) _ = Left ""
append _ (Pure _) = Left "" 
append (Free xs) (Free ys) = fmap Free $ sequenceA $ zipWith append xs ys

此定义与您给出的定义同构。

于 2014-01-17T03:47:38.407 回答