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为了帮助我学习 Applicative Functors 和 Functors，我认为看看如何Either使用 typeclassesFunctor和Applicative. 显然我可以继续阅读代码，但我发现自己尝试实现事物以更好地理解事物更有用。

仅供参考，我正在尝试实现此演示文稿结果的 Haskell 版本http://applicative-errors-scala.googlecode.com/svn/artifacts/0.6/chunk-html/index.html

无论如何，这就是我到目前为止所拥有的

 data Validation a b = Success a | Failure b deriving (Show, Eq)

 instance Functor (Validation a) where
     fmap f (Failure x) = Failure x
     fmap f (Success x) = Success (f x)

但是每当我尝试运行它时，ghci我都会收到以下错误消息：-

[1 of 1] Compiling Main             ( t.hs, interpreted )

t.hs:5:35:
    Couldn't match type `b' with `a1'
      `b' is a rigid type variable bound by
          the type signature for
            fmap :: (a1 -> b) -> Validation a a1 -> Validation a b
          at t.hs:4:5
      `a1' is a rigid type variable bound by
           the type signature for
             fmap :: (a1 -> b) -> Validation a a1 -> Validation a b
           at t.hs:4:5
    Expected type: a
      Actual type: b
    In the return type of a call of `f'
    In the first argument of `Success', namely `(f x)'
    In the expression: Success (f x)

t.hs:5:37:
    Couldn't match type `a' with `a1'
      `a' is a rigid type variable bound by
          the instance declaration at t.hs:3:30
      `a1' is a rigid type variable bound by
           the type signature for
             fmap :: (a1 -> b) -> Validation a a1 -> Validation a b
           at t.hs:4:5
    In the first argument of `f', namely `x'
    In the first argument of `Success', namely `(f x)'
    In the expression: Success

我不确定这是为什么，有人可以帮忙吗？

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1 回答 1

13

您正在尝试使Functor实例在Success零件上工作，这是正常的做法，但是由于您的类型参数的顺序，它是在Failure零件中的类型上定义的。

由于您已将其定义为

data Validation a b = Success a | Failure b

instance Functor (Validation a) where
    ...

这意味着您的实现fmap应该具有类型(x -> y) -> Validation a x -> Validation a y。但是由于第二个类型变量是针对这种Failure情况的，所以这不会进行类型检查。

您希望Success案例的类型变量是最后一个：

data Validation b a = Success a | Failure b
于 2011-08-19T12:08:55.577 回答