haskell - 可以部分应用仿函数吗

Question

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？ ^

Answer 1

Does it mean that all a-s within the Tree must somehow become b-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 Node::b -> Tree b -> Tree b -> Tree b::树a，以及右::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。