Question

我已经编写了这段代码来打印一张树：

data Node = A | B | C | Tree Node [Node]
    deriving (Show, Eq)

printTree' :: Int -> Node -> [String]
printTree' spaceCount (Tree node children) = (printTree' spaceCount node)
    ++ concat (map (\child -> printTree' (spaceCount+2) child) children)
printTree' spaceCount leaf = [(replicate spaceCount ' ') ++ (show leaf)]

printTree :: Node -> String
printTree node = unlines (printTree' 0 node)

示例输出：

*Main> putStr $ printTree $ Tree A [Tree A [A,B,C], C]
A
  A
    A
    B
    C
  C

现在我想将此作为show的实现。这种方法很接近，但我找不到一种方法来调用内置的show：

instance Show Node where
    show (Tree node children) = printTree (Tree node children)
    show _ = "node... can I call the built-in show here?"

（在这个例子中，我可以处理A，B和C.但是在实际代码中，有许多节点类型。）

Answer 1

我能看到的唯一方法就是分成两种类型。

data Tree node = Tree node [Tree node]
data Node = A | B | C deriving Show

instance Show node => Show (Tree node) where ....

Answer 2

遵循MathematicalOrchid的回应，最好的方法是使用新类型，但这是组织类型的更好方法：

data Node = Leaf Leaf | Tree Node [Node] deriving Eq
data Leaf = A | B | C deriving (Eq, Show)

instance Show Node where
  show (Tree node children) = printTree (Tree node children)
  show (Leaf l) = show l

我可以部分派出秀吗？

2 个答案: