haskell - 没有因使用 ‘==’ 而产生 (Eq a) 的实例

Question

我正在尝试为此数据类型编写一个 Eq 实例:

data Tree a b = Leaf b | Node a (Tree a b) (Tree a b)
deriving (Show)

我编写了我认为可行的简单代码:

instance Eq (Tree a b) where
  (Leaf x) == (Leaf y) =  x == y
  (Node val1 l1 r1) == (Node val2 l2 r2)  =  (val1 == val2) && (l1==l2) && (r1==r2)
  _  == _  = False

但随后我收到错误:

• No instance for (Eq a) arising from a use of ‘==’
  Possible fix: add (Eq a) to the context of the instance declaration
• In the first argument of ‘(&&)’, namely ‘(val1 == val2)’
  In the expression: (val1 == val2) && (l1 == l2) && (r1 == r2)
  In an equation for ‘==’:
     (Node val1 l1 r1) == (Node val2 l2 r2)
     = (val1 == val2) && (l1 == l2) && (r1 == r2)

我尝试添加 Eq a => ... 但随后类型 b 出现相同的错误。我似乎也无法添加 Eq b 。

任何帮助将不胜感激，10x!

Answer 1

正如您所写，您必须设置类型约束Eq a 和Eq b。您只需将约束放在括号中并用逗号分隔即可。

instance (Eq a, Eq b) => Eq (Tree a b) where
  (Leaf x) == (Leaf y) =  x == y
  (Node val1 l1 r1) == (Node val2 l2 r2)  =  (val1 == val2) && (l1==l2) && (r1==r2)
  _  == _  = False