部分应用函数类型的函子 [英] functors from partially applied function type

问题描述

在Haskell编程中有一个问题说:完成以下声明:

there is a quesion in Programming in Haskell that says: Complete the following declaration:

instance Functor ((->) a) where

现在作为Functor Thing的类型定义为:

Now as Functor Thing has a type definition of:

instance Functor Thing where    
   --fmap::(a -> b) -> Thing a -> Thing b 

我想知道这种减少是否有意义:

I was wondering if this reduction makes sense:

instance Functor ((->) a) where
    -- fmap::(a -> b) -> ((->) a) a -> ((->) a) b 
    -- therefore
    -- fmap::(a -> b) -> a -> a -> (a -> b) 
    -- therefore
    -- fmap::b -> b

-更新--- 我错过了括号，应该是

-- update --- I missed brackets, it should have been

instance Functor ((->) a) where
    -- fmap::(a -> b) -> ((->) a) a -> ((->) a) b 
    -- therefore
    -- fmap::(a -> b) -> (a -> a) -> (a -> b) 
    -- therefore
    -- I should be returning a function of a -> b

推荐答案

否，因为实例声明中的 a 与中的 a 不同 fmap 的类型.您需要在实例声明中分配一个类型变量，以避免在捕获" fmap 的类型:

No, because the a in your instance declaration is not the same a as the one in the type of fmap. You need to assign a type variable in your instance declaration that avoids "capturing" the a in the type of fmap:

instance Functor ((->) r) where
  fmap :: (a -> b) -> (r -> a) -> (r -> b)

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