haskell - 我可以部分导出 Show 吗？

Question

我编写了这段代码来漂亮地打印一棵树:

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

现在我想将其作为 show 的实现。这种方法很接近，但我找不到调用内置 show 的方法:

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，我可以只处理A、B和C。但在实际代码中，有很多节点类型。)

Answer 1

我认为做到这一点的唯一方法是将其分为两种类型。

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

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