使用'show'时没有（Show a）的实例

Question

使用'show'时没有（Show a）的实例在'（++）'的第一个参数中，即'show a'

data LTree a = Leaf a | Node (LTree a) (LTree a)
instance Show (LTree a) where
    show (Leaf a) = "{" ++ show a ++ "}"
    show (Node fe fd) = "<" ++ (show fe)++ "," ++(show fd)++ ">"

Node (Leaf 1) (Node (Node (Leaf 3) (Leaf 4)) (Node (Leaf 8) (Leaf 7)))

我应该得到：

<{1},<<{3},{4}>,<{8},{7}>>>

Answer 1

在你的行：

    show (Leaf a) = "{" ++ show a ++ "}"

你调用show a，a是a类型的元素，但是并不是说a类型是Show的一个实例，因此你需要为你的instance声明添加一个约束：

instance Show a => Show (LTree a) where
    show (Leaf a) = "{" ++ show a ++ "}"
    show (Node fe fd) = "<" ++ (show fe)++ "," ++(show fd)++ ">"

所以在这里我们说LTree a是一个节目的实例给定a是Show的一个实例。对于您给定的样本数据，我们获得：

Prelude> Node (Leaf 1) (Node (Node (Leaf 3) (Leaf 4)) (Node (Leaf 8) (Leaf 7)))
<{1},<<{3},{4}>,<{8},{7}>>>