如何写地图/ fmap模拟（a-> b） - > F m a - > F m b

Question

好日/夜大家！ 我有型：

data FixItem m a = KeepItem|SkipItem|FixItem (m (a -> a)) 
fixItem f = FixItem $ pure f 

，我想写功能mapFix :: (a -> b) -> FixItem m a -> FixItem m b。当我尝试：

mapFix f SkipItem = SkipItem -- good 
mapFix f KeepItem = fixItem f -- error "rigid type"!!! 
mapFix f (FixItem mf) = FixItem $ pure (.) <*> (pure f) <*> mf -- too! 

所以，我得到错误：

• Couldn't match type ‘b’ with ‘a’ 
     ‘b’ is a rigid type variable bound by 
     the type signature for: 
      mapFix :: forall (m :: * -> *) a b. 
        Applicative m => 
        (a -> b) -> FixItem m a -> FixItem m b 
     at src/test.hs:235:11 
     ‘a’ is a rigid type variable bound by 
     the type signature for: 
      mapFix :: forall (m :: * -> *) a b. 
        Applicative m => 
        (a -> b) -> FixItem m a -> FixItem m b 
     at src/test.hs:235:11 
     Expected type: b -> b 
     Actual type: a -> b 
    • In the first argument of ‘fixItem’, namely ‘f’ 
     In the expression: fixItem f 
     In an equation for ‘mapFix’: mapFix f KeepItem = fixItem f 
    • Relevant bindings include 
     f :: a -> b (bound at src/test.hs:236:8) 
     mapFix :: (a -> b) -> FixItem m a -> FixItem m b 
      (bound at src/test.hs:236:1) 

如何写mapFix或实现函子实例这种类型（FixItem修复a到a，不b，IE修复是a -> a ，而不是a -> b）？

Answer 1

您不能为您的数据类型实施Functor类型类。这是因为你的一个构造函数中有 a -> a。当你有功能时，你应该更加小心。但简而言之，您将逆变类型变量a键入，因此您无法在此类型变量上实现Functor。

尽管您可以为您的数据类型实施Invariant。因为a在协变和逆变职位，您的数据类型是不变的函子。

可以帮助您：

Example of Invariant Functor?

What is a contravariant functor?

Some blog post