Question

我是Haskell的新手并且一直在尝试构建一个压缩函数，该函数适用于具有以下数据结构的树：

data Tree a = Leaf | Node a (Tree a) (Tree a) deriving Show

到目前为止，我有这个：

treezip :: (Tree a) -> (Tree b) -> (Tree(a,b))
treezip (Node a leftSubtreea rightSubtreea) (Node b leftSubtreeb rightSubtreeb) =
let l = treezip leftSubtreea leftSubtreeb
    r = treezip rightSubtreea rightSubtreeb
in Node a l r

然而，每当我尝试将模块加载到GHCi中时，我都会收到错误，指向最后一行代码并特别关注变量a。

我一直绞尽脑汁想弄清楚为什么这不起作用。任何帮助将不胜感激

Answer 1

不要对编译器起作用，让它帮助你！

为了使这样的函数正确，最好只从“外壳定义”开始：

treeZip (Node a lSa rSa) (Node b lSb rSb) = Node _ _ _

编译器会回来：

• Found hole: _ :: (a, b)
  Where: ‘a’ is a rigid type variable ...(_bla, bla_)
         ‘b’ is a rigid type variable ....
• In the first argument of ‘Node’, namely ‘_’
  In the expression: Node _ _ _
  In an equation for ‘treeZip’:
      treeZip (Node a lSa rSa) (Node b lSb rSb) = Node _ _ _
• Relevant bindings include
    rSb :: Tree b (bound at /tmp/wtmpf-file20584.hs:4:38)
    lSb :: Tree b (bound at /tmp/wtmpf-file20584.hs:4:34)
    b :: b (bound at /tmp/wtmpf-file20584.hs:4:32)
    rSa :: Tree a (bound at /tmp/wtmpf-file20584.hs:4:21)
    lSa :: Tree a (bound at /tmp/wtmpf-file20584.hs:4:17)
    a :: a (bound at /tmp/wtmpf-file20584.hs:4:15)

因此，它会告诉您第一个_ 类型的孔应填充(a,b)类型的内容。我们有这种类型的东西吗？好吧，我们可以轻松地构建它，即通过a和b并将它们放在一个元组中！

treezip (Node a lSa rSa) (Node b lSb rSb) = Node (a,b) _ _

...给

• Found hole: _ :: Tree (a, b)
  Where: ‘a’ is a rigid type variable bound by...

是的，你已经解决了那个 - 它应该是子树的拉链。

treeZip (Node a lSa rSa) (Node b lSb rSb)
     = Node (a,b) (treeZip lSa lSb) (treeZip rSa rSb)

Haskell：拉树的问题

1 个答案: