为什么向工作函数添加as-pattern会导致编译错误？

Question

Here's Either a的标准函数器实例

instance Functor (Either a) where
        fmap _ (Left x) = Left x
        fmap f (Right y) = Right (f y)

当加载到GHCi中时，添加as-pattern会导致编译错误：

instance Functor (Either a) where
        fmap _ z@(Left x) = z          -- <-- here's the as-pattern
        fmap f (Right y) = Right (f y)

Couldn't match expected type `b' against inferred type `a1'
  `b' is a rigid type variable bound by
      the type signature for `fmap' at <no location info>
  `a1' is a rigid type variable bound by
       the type signature for `fmap' at <no location info>
  Expected type: Either a b
  Inferred type: Either a a1
In the expression: z
In the definition of `fmap': fmap _ (z@(Left x)) = z

为什么这个不起作用？

Answer 1

fmap具有签名(a -> b) -> f a -> f b，即它必须允许a和b不同。在您的实现中，a和b只能相同，因为它返回的内容与作为参数传递的内容相同。所以GHC抱怨说。

Answer 2

我最好的猜测是这是失败的，因为z在等式的每一端代表不同的类型：

整体类型为fmap :: (a -> b) -> Either t a -> Either t b

• on

• left，z :: Either t b

，

似乎允许Left x在同一等式中有多个不同的类型，但z不允许。

这个实现也失败了，显然是出于同样的原因：

instance Functor (Either a) where
        fmap f (Right y) = Right (f y)
        fmap _ z = z          

Couldn't match expected type `b' against inferred type `a1'
  `b' is a rigid type variable bound by
      the type signature for `fmap' at <no location info>
  `a1' is a rigid type variable bound by
       the type signature for `fmap' at <no location info>
  Expected type: Either a b
  Inferred type: Either a a1
In the expression: z
In the definition of `fmap': fmap _ z = z

Answer 3

如果您将fmap的签名专门化为Either l，则会得到：

fmap :: (a -> b) -> Either l a -> Either l b

这意味着在case语句左侧进行模式匹配的Left r必须具有Either l a类型。但是，您不能按原样返回它，因为您必须返回Either l b。这需要将左边的值重新包装在一个新的Left中，以便编译器可以推断它返回的是一个新生成的Either，该Right值可能具有不同的类型。