多个层次的函子

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：我在那里做错了什么？

Answer 1

当你陷入这样的困境时，我建议使用类型孔，因为它们会引导你走向正确的方向

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fmap2 :: (Functor f, Functor g) => (a -> b) -> f (g a) -> f (g b)\nfmap2 f nested_functor = fmap _something nested_functor \n--      |                |    |- notice this underscore\n--      |                |- clearly we need to fmap something over the nested_functor but we don\'t know what yet\n--      |- Using better names usually helps. Haskellers tend to use very short names, but that\'s not appropriate always\n

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如果您尝试编译此代码，ghc将会注意到_something并会建议您使用它的类型

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Found hole: _something :: g a -> g b <-- This is interesting\n      Where: \xe2\x80\x98a\xe2\x80\x99, \xe2\x80\x98g\xe2\x80\x99, \xe2\x80\x98b\xe2\x80\x99 ... (blah blah blah, not interesting)\n      Or perhaps \xe2\x80\x98_something\xe2\x80\x99 is mis-spelled, or not in scope\n

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我们知道需要给出_something类型。因此：g a -> g bFunctor g

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Found hole: _something :: g a -> g b <-- This is interesting\n      Where: \xe2\x80\x98a\xe2\x80\x99, \xe2\x80\x98g\xe2\x80\x99, \xe2\x80\x98b\xe2\x80\x99 ... (blah blah blah, not interesting)\n      Or perhaps \xe2\x80\x98_something\xe2\x80\x99 is mis-spelled, or not in scope\n

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