如何编写 map/fmap 模拟 (a->b) -> F m a -> F m b

Question

大家早安/晚安！ 我有类型：

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

我想写函数mapFix :: (a -> b) -> FixItem m a -> FixItem m b。当我尝试时：

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)

如何为这种类型编写 mapFix 或实现 Functor 实例（FixItem 将a 修复为a，而不是b，即修复为a -> a，而不是a -> b）？

Answer 1

您不能为您的数据类型实现Functor 类型类。这是因为 a -> a 在您的一个构造函数中。当你有函数时，你应该更加小心。但简而言之，你有类型变量 a 在逆变位置，所以你不能在这个类型变量上实现 Functor。

虽然您可以为您的数据类型实现Invariant。因为a 在 covariant 和 contravariant 位置，您的数据类型是不变函子。

可以帮助你：

Example of Invariant Functor?

What is a contravariant functor?

Some blog post