（show a）的任何实例均不因使用“ show”而产生

Question

在（（++））的第一个参数中，“ show a”的使用不会导致（Show a）的实例。

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 ++ "}"

您可以show a使用a类型为的元素调用，a但是并不能说该类型a是的实例Show，因此需要在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，这里我们说的LTree a是show的一个实例，而给定的 a是的一个实例Show。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}>>>