为什么在工作函数中添加as-pattern会导致编译错误？

Question

这是标准的Functor实例Either a:

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

在加载到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不同.在您的实现中,a并且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抱怨道.