我正在尝试为以下类型实现fmap
:
data Tree a = Leaf a | Node a (Tree a) (Tree a) | Empty deriving (Eq,Show)
instance Functor Tree where
fmap _ Empty=Empty
fmap f (Leaf x)=Leaf (f x)
fmap f (Node t left right)=Node (f t) left right
我不断收到类型不匹配错误:
错误
* Couldn't match type `a' with `b'
`a' is a rigid type variable bound by
the type signature for:
fmap :: forall a b. (a -> b) -> Tree a -> Tree b
at Monad.hs:8:9-12
`b' is a rigid type variable bound by
the type signature for:
fmap :: forall a b. (a -> b) -> Tree a -> Tree b
at Monad.hs:8:9-12
Expected type: Tree b
Actual type: Tree a
为什么会出现此错误,但是当我还将fmap
应用于子节点时,它也会毫无问题地进行编译:
fmap f (Node t left right) = Node (f t) (fmap f left) (fmap f right)
这是否意味着a
内的所有Tree
-s必须以某种方式变为b
-s?我只在第一种情况下与非功能者打交道?
^
答案 0 :(得分:8)
这是否意味着
a
内的所有Tree
-s必须以某种方式变为b
-s?我只在第一种情况下与非功能者打交道? ^
是的,没错。您正在尝试实现fmap :: (a -> b) -> Tree a -> Tree b
,但是在编写时:
fmap f (Node t left right) = Node (f t) left right
您正在尝试使用参数Node :: b -> Tree b -> Tree b -> Tree b
,f t :: b
和left :: Tree a
来调用right :: Tree a
。将Tree a
转换为Tree b
的唯一方法是通过fmap f :: Tree a -> Tree b
,这就是为什么:
fmap f (Node t left right) = Node (f t) (fmap f left) (fmap f right)
按预期工作。