Question

代码：

connected edges = connect slist tslist edges
    where tslist = tail(slist)
          slist = sort (remdup (myflatten edges))


connect _ [] _ = True        
connect (x:xs) slist edges = if (findpaths x slist edges) then (connect xs list edges) else False
    where list = tail(slist) 

findpaths _ [] _ = True
findpaths x (y:ys) edges = if (findpath x y edges edges []) then (findpaths x ys edges)
                       else False

findpath _ _ [] _ _ = False                         
findpath x y ((w,z):xs) edges nodes | ((x==w && y==z) || (x==z && y==w)) = True  
line 57                             | (x==w && not(member z nodes)) = (findpath z y edges edges nodes++[z] || findpath y z edges edges nodes++[z])
line 58                             | (x==z && not(member w nodes)) = (findpath w y edges edges nodes++[w] || findpath y w edges edges nodes++[w])
                                    | otherwise = findpath x y xs edges nodes

“少数”错误：

project2.hs:57:72:
Couldn't match expected type `[a0]' with actual type `Bool'
In the return type of a call of `findpath'
In the first argument of `(++)', namely
  `findpath z y edges edges nodes'
In the first argument of `(||)', namely
  `findpath z y edges edges nodes ++ [z]'

project2.hs:57:72:
Couldn't match expected type `Bool' with actual type `[a0]'
In the first argument of `(||)', namely
  `findpath z y edges edges nodes ++ [z]'
In the expression:

  (findpath z y edges edges nodes ++ [z]
   || findpath y z edges edges nodes ++ [z])
In an equation for `findpath':
    findpath x y ((w, z) : xs) edges nodes
      | ((x == w && y == z) || (x == z && y == w)) = True
      | (x == w && not (member z nodes))
      = (findpath z y edges edges nodes ++ [z]
         || findpath y z edges edges nodes ++ [z])
      | (x == z && not (member w nodes))
      = (findpath w y edges edges nodes ++ [w]
         || findpath y w edges edges nodes ++ [w])
      | otherwise = findpath x y xs edges nodes

project2.hs:57:111:
Couldn't match expected type `[a0]' with actual type `Bool'
In the return type of a call of `findpath'
In the first argument of `(++)', namely
  `findpath y z edges edges nodes'
In the second argument of `(||)', namely
  `findpath y z edges edges nodes ++ [z]'

project2.hs:57:111:
Couldn't match expected type `Bool' with actual type `[a0]'
In the second argument of `(||)', namely
  `findpath y z edges edges nodes ++ [z]'
In the expression:
  (findpath z y edges edges nodes ++ [z]
   || findpath y z edges edges nodes ++ [z])
In an equation for `findpath':
    findpath x y ((w, z) : xs) edges nodes
      | ((x == w && y == z) || (x == z && y == w)) = True
      | (x == w && not (member z nodes))
      = (findpath z y edges edges nodes ++ [z]
         || findpath y z edges edges nodes ++ [z])
      | (x == z && not (member w nodes))
      = (findpath w y edges edges nodes ++ [w]
         || findpath y w edges edges nodes ++ [w])
      | otherwise = findpath x y xs edges nodes

project2.hs:58:72:
Couldn't match expected type `[a0]' with actual type `Bool'
In the return type of a call of `findpath'
In the first argument of `(++)', namely
  `findpath w y edges edges nodes'
In the first argument of `(||)', namely
  `findpath w y edges edges nodes ++ [w]'

project2.hs:58:72:
Couldn't match expected type `Bool' with actual type `[a0]'
In the first argument of `(||)', namely
  `findpath w y edges edges nodes ++ [w]'
In the expression:
  (findpath w y edges edges nodes ++ [w]
   || findpath y w edges edges nodes ++ [w])
In an equation for `findpath':
    findpath x y ((w, z) : xs) edges nodes
      | ((x == w && y == z) || (x == z && y == w)) = True
      | (x == w && not (member z nodes))
      = (findpath z y edges edges nodes ++ [z]
         || findpath y z edges edges nodes ++ [z])
      | (x == z && not (member w nodes))
      = (findpath w y edges edges nodes ++ [w]
         || findpath y w edges edges nodes ++ [w])
      | otherwise = findpath x y xs edges nodes

project2.hs:58:111:
Couldn't match expected type `[a0]' with actual type `Bool'
In the return type of a call of `findpath'
In the first argument of `(++)', namely
  `findpath y w edges edges nodes'
In the second argument of `(||)', namely
  `findpath y w edges edges nodes ++ [w]'

project2.hs:58:111:
Couldn't match expected type `Bool' with actual type `[a0]'
In the second argument of `(||)', namely
  `findpath y w edges edges nodes ++ [w]'
In the expression:
  (findpath w y edges edges nodes ++ [w]
   || findpath y w edges edges nodes ++ [w])
In an equation for `findpath':
    findpath x y ((w, z) : xs) edges nodes
      | ((x == w && y == z) || (x == z && y == w)) = True
      | (x == w && not (member z nodes))
      = (findpath z y edges edges nodes ++ [z]
         || findpath y z edges edges nodes ++ [z])
      | (x == z && not (member w nodes))
      = (findpath w y edges edges nodes ++ [w]
         || findpath y w edges edges nodes ++ [w])
      | otherwise = findpath x y xs edges nodes

有人可以为了上帝的缘故向我解释这些f ***匹配是如何运作的吗？我做了一些更改，但我改变了正确的事情，无法找到问题的根源。我是haskell的新手，PL很奇怪，因为我至少认为，没有严格的功能格式化，匹配过程让我感到困惑。谢谢你。

Answer 1

你需要围绕（节点++ [z]）

另外，它将像这样计算

(findpath z y edges edges nodes)++[z]

函数应用程序总是比任何中缀操作符更强大

haskell中的多个匹配错误

1 个答案: