haskell - 双类型仿函数定义被拒绝

Question

为什么这个仿函数定义被拒绝？

data Second a b = Ok a b | Ko a b deriving (Show)

instance Functor (Second x) where
  fmap f (Ok a b ) = Ok (f a) b
  fmap f (Ko a b ) = Ko a (f b) 

我 get a lot of errors :

GHCi, version 8.0.1

main.hs:4:22: error:
    • Couldn't match type ‘b’ with ‘x’
      ‘b’ is a rigid type variable bound by
        the type signature for:
          fmap :: forall a b. (a -> b) -> Second x a -> Second x b
        at main.hs:4:3
      ‘x’ is a rigid type variable bound by
        the instance declaration at main.hs:3:10
      Expected type: Second x b
        Actual type: Second b a
    • In the expression: Ok (f a) b
      In an equation for ‘fmap’: fmap f (Ok a b) = Ok (f a) b
      In the instance declaration for ‘Functor (Second x)’
    • Relevant bindings include
        a :: x (bound at main.hs:4:14)
        f :: a -> b (bound at main.hs:4:8)
        fmap :: (a -> b) -> Second x a -> Second x b (bound at main.hs:4:3)
main.hs:4:28: error:
    • Couldn't match expected type ‘a’ with actual type ‘x’
      ‘x’ is a rigid type variable bound by
        the instance declaration at main.hs:3:10
      ‘a’ is a rigid type variable bound by
        the type signature for:
          fmap :: forall a b. (a -> b) -> Second x a -> Second x b
        at main.hs:4:3
    • In the first argument of ‘f’, namely ‘a’
      In the first argument of ‘Ok’, namely ‘(f a)’
      In the expression: Ok (f a) b
    • Relevant bindings include
        b :: a (bound at main.hs:4:16)
        a :: x (bound at main.hs:4:14)
        f :: a -> b (bound at main.hs:4:8)
        fmap :: (a -> b) -> Second x a -> Second x b (bound at main.hs:4:3)

这是什么意思呢？请帮忙。

Answer 1

如果你解开 Functor 的定义你会看到 Second的第一个参数保持不变，而它的第二个参数用于转换。

class Functor f where
  fmap :: (s -> t) -> f s -> f t

(我已将类型变量 a 和 b 分别重命名为 s 和 t ，因为您还有值变量 a 和 b 。)

那 class给我们，在您的具体 instance

instance Functor (Second x) where
  -- fmap :: (s -> t) -> (Second x) s -> (Second x) t
  --    i.e. (s -> t) -> Second x s   -> Second x t

现在，当你实现 fmap ，您必须确保 fmap 的用户可以任意选择 x , s和 t他们要。所以你不能对它们做任何假设:它们代表任意的，可能是不同的类型。这就是错误消息在谈论“刚性”类型变量时的含义:您的代码不允许为它们选择特定类型，以便其用户可以选择。编译器提示您 promise 提供一个非常多态的函数，但提供了一个较少多态的函数，该函数仅在 x=s=t 时才进行类型检查。 .

也就是说，当你写

  fmap f (Ok a b) = Ok (f a) b

你有

f :: s -> t
Ok a b :: Second x s
a :: x
b :: s

你回来了

Ok (f a) b :: Second x t

这需要

f a :: x   -- clearly not true, as f :: s -> t
b :: t     -- clearly not true, as b :: s

在哪里 f a需要

a :: s     -- clearly not true, as a :: x

所以，是的，很多错误。

一个 Functor instance 允许您在与类型的最后一个参数对应的位置转换数据，即 b在

data Second a b = Ok a b | Ko a b deriving (Show)

所以你的

  fmap f (Ko a b ) = Ko a (f b)

很好，但是你的

  fmap f (Ok a b ) = Ok (f a) b

命中了不同参数的使用并且不正常。

  fmap f (Ok a b ) = Ok a (f b)

会工作。与更改 data 一样声明

data Second a b = Ok b a | Ko a b deriving (Show)

并留下您的 instance照原样。

任何一种修复都有效，但不要同时“为了安全起见”!