haskell - 我无法为我的 Forest 数据类型使用这个 Show/Functor 实例

Question

我大约两个月前就开始用 Haskell 编程了，我还很新，所以不要指望我能成为 monad 或其他方面的顶级人物。我尝试了很多方法来让这个 Forest 数据类型成为仿函数的实例并显示，但我真的不知道如何解决编译器给我的冲突。如:

    Not in scope: data constructor ‘Tree’
    Perhaps you meant ‘True’ (imported from Prelude)
   |
15 |     show ((Tree a) : (Forest s) ) = "[" ++ show a ++ "," ++ show s ++ "]"
   |            ^^^^

exercici3.hs:15:23: error: Not in scope: data constructor ‘Forest’
   |
15 |     show ((Tree a) : (Forest s) ) = "[" ++ show a ++ "," ++ show s ++ "]"
   |                       ^^^^^^

exercici3.hs:19:12: error:
    Not in scope: data constructor ‘Tree’
    Perhaps you meant ‘True’ (imported from Prelude)
   |
19 |     fmap ((Tree a) : (Forest s)) = [f a] ++ (fmap f s)
   |            ^^^^

exercici3.hs:19:23: error: Not in scope: data constructor ‘Forest’
   |
19 |     fmap ((Tree a) : (Forest s)) = [f a] ++ (fmap f s)
   |                       ^^^^^^

这是类的字体代码。想了很久没有找到合理的解决办法，欢迎大家帮忙，谢谢!

data Tree a = Empty | Node a (Tree a) (Tree a)
data Forest a = Nil  | Cons (Tree a) (Forest a)

instance Show a => Show (Tree a) where
    show Empty = "()"
    show (Node b (xl) (xr)) = "(" ++ show xl ++ "," ++ (show b) ++  "," ++ show xr ++ ")"

instance Functor (Tree ) where
    fmap  f Empty  = Empty
    fmap  f (Node a (xl) (xr)) = Node (f a) (fmap f xl) (fmap f xr)

instance Show a => Show (Forest a) where
    show Nil = []
    show ((Tree a) : (Forest s) ) = "[" ++ show a ++ "," ++ show s ++ "]"

instance Functor (Forest) where
    fmap f Nil = []
    fmap ((Tree a) : (Forest s)) = [f a] ++ (fmap f s)

需要明确的是，Tree 数据类型工作得很好，只是森林的语法部分似乎根本不起作用。

Answer 1

数据构造函数是Cons，而不是(:)。然后，您使用例如 x 和 xs 作为变量，其中 x 的类型为 Tree a，并且 xs 的类型为 Forest a:

instance Show a => Show (Forest a) where
    show Nil = ""
    show (Cons x xs) = "[" ++ show x ++ "," ++ show xs ++ "]"

instance Functor Forest where
    fmap f Nil = Nil
    fmap f (Cons x xs) = Cons (fmap f x) (fmap f xs)

话虽如此，我认为没有太多理由在这里定义数据类型 Forest，您可以将其定义为:

type Forest a = [Tree a]