haskell - 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 Functor 实例:

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

在加载到 GHCi 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

为什么这不起作用？

Answer 1

fmap有签名(a -> b) -> f a -> f b ，即它必须允许 a和 b与众不同。在您的实现中，a和 b只能是相同的，因为它返回作为参数传递的相同内容。所以GHC提示。