Haskell自定义类型Show实例问题：构造器不在作用域及调用报错

Let's walk through your problems step by step:

1. The treeFromList [1,2,3] Runtime Error #

The error you're seeing happens because of how treeFromList is defined. Its type signature is:

treeFromList :: (Ord a) => [a] -> BTree a -> BTree a

This means it expects two arguments: a list of values, and an initial tree to insert into. When you call treeFromList [1,2,3] without the second argument, Haskell returns a function of type BTree Integer -> BTree Integer instead of a concrete tree. Functions don't have a default Show instance, which triggers the error when you try to print it.

Fixes: #

• Option 1: Pass an initial empty tree explicitly:

  treeFromList [1,2,3] EmptyTree
  

• Option 2: Refactor treeFromList to start with an empty tree by default (this is more idiomatic):

  treeFromList :: (Ord a) => [a] -> BTree a
  treeFromList = foldr treeInsert EmptyTree
  

  Now you can call treeFromList [1,2,3] directly, and it will return the built tree which can be printed (thanks to your deriving Show clause).

2. Custom Show Instance Compilation Errors #

Your attempt to define custom Show instances has a key misunderstanding: EmptyTree and Node are data constructors for the BTree a type, not separate types. You can't define Show instances for them individually—you need to define a single instance for the BTree a type, pattern-matching on its constructors.

Here's why your code failed:

• You tried to pattern-match on BTree a, but BTree is a type constructor (used to create types like BTree Integer), not a data constructor (used to create values).
• You tried to create instances for EmptyTree and Node, which aren't standalone types.

Correct Custom Show Instance Example #

If you want a custom output format (instead of the auto-generated one from deriving Show), here's a valid implementation:

instance Show a => Show (BTree a) where
    show EmptyTree = "EmptyTree"
    show (Node left root right) = "Node (" ++ show left ++ ") " ++ show root ++ " (" ++ show right ++ ")"

This will produce output like:

Node (Node (EmptyTree) 1 (EmptyTree)) 2 (Node (EmptyTree) 3 (EmptyTree))

If you prefer a simpler format (like in-order traversal), you could write:

instance Show a => Show (BTree a) where
    show EmptyTree = ""
    show (Node left root right) = show left ++ " " ++ show root ++ " " ++ show right

Which would output:

1  2  3 

Remember: since you already used deriving Show on your BTree type, you only need a custom instance if you want to change the default output format.

内容的提问来源于stack exchange，提问作者twigonometry