Question

（显示a）的任何实例均不因使用“显示”而产生在“（++）”的第一个参数中，即“显示一个”

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

在您的行中：

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

您调用show a，其中a是类型a的元素，但不是表示该类型a是一个实例Show的值，因此您需要在instance声明中添加约束：

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

因此，我们在这里说LTree a是show give 的实例。a是Show的实例。对于您给定的样本数据，我们将获得：

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

没有因使用“显示”而引起的（显示a）实例

1 个答案: