没有因使用“show”而产生的（Show a）实例

Question

没有因使用“show”而产生的 (Show a) 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

在你的行中：

    show (Leaf a) = "{" ++ <b>show a</b> ++ "}"

你调用show a，用a一个类型a的元素，但是不是说那个类型a 是 Show 的实例，因此您需要在 instance 声明中添加约束：

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

所以这里我们说LTree a是show的一个实例givena是Show的一个实例。对于您给定的示例数据，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“show”而产生的（Show a）实例

No instance for (Show a) arising from a use of ‘show’

haskell

show