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
为什么我会收到此错误,但当我也将 fmap
应用于子节点时,它编译时没有问题:
fmap f (Node t left right) = Node (f t) (fmap f left) (fmap f right)
这是否意味着 Tree
中的所有 a
-s 必须以某种方式变成 b
-s ?我只处理第一种情况中的非仿函数?
^
最佳答案
Does it mean that all
a
-s within theTree
must somehow becomeb
-s ? and i am only dealing with the non-functor one in the first case ? ^
是的,没错。您正在尝试实现 fmap::(a -> b) -> Tree a -> Tree b
,但是当您编写时:
fmap f (Node t left right) = Node (f t) left right
您正在尝试使用参数 f t::b
, left 调用
,以及Node::b -> Tree b -> Tree b -> Tree b
::树a右::树a
。将 Tree a
转换为 Tree b
的唯一方法是通过 fmap f::Tree a -> Tree b
,即为什么这样:
fmap f (Node t left right) = Node (f t) (fmap f left) (fmap f right)
按预期工作。
关于haskell - 可以部分应用仿函数吗,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55701842/