多层次函子

Question

我有这个蹩脚的尝试：

fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)
fmap2 f f2 = (fmap2 f . fmap f2)

它应该像这样工作：

fmap2 negate [[1,2], [3]] -- Returns [[-1,-2],[-3]]
fmap2 head [Just "abc",Nothing,Just "def"] -- Returns [Just 'a',Nothing,Just 'e']

请注意，这是一个练习。 （我尝试了大约1.5个小时，但没有找到任何解决方案。）

问：我做错了什么？

Answer 1

当你陷入这样的困境时。我建议使用类型孔，因为它们会带您朝正确的方向前进

fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)
fmap2 f nested_functor = fmap _something nested_functor 
--      |                |    |- notice this underscore
--      |                |- clearly we need fo fmap something over the nested_functor but we don't know what yet
--      |- Use better names usually help. Haskellers tend to use ver short names, but that's not appropiate always

如果您尝试编译此代码，

ghc

会注意到

_something

，并会建议您使用它的类型

Found hole: _something :: g a -> g b <-- This is interesting
      Where: ‘a’, ‘g’, ‘b’ ... (blah blah blah, not interesting)
      Or perhaps ‘_something’ is mis-spelled, or not in scope

我们知道，给定

_something

，

g a -> g b

需要具有类型

Functor g

。因此：

fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)
fmap2 f nested_functor = fmap something nested_functor
  where 
    -- remember something :: Functor g => g a -> g b
    something = undefined -- left as an exercise

Answer 2

您应该

fmap

使用函子映射作为内部结构的函数，所以类似于：

fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)
fmap2 f = fmap (fmap …)

我留下填写

…

部分作为练习。