haskell - fmap的函数组成

Question

函数组合的简单定义是:

f ( g x)

或

(f . g) $ x

现在我有以下示例:

  newtype Compose f g a =
    Compose { getCompose :: f (g a) }
    deriving (Eq, Show)

  instance (Functor f, Functor g) => Functor (Compose f g) where
    fmap f (Compose fga) = Compose $ (fmap . fmap) f fga

然后我尝试将不使用组合运算符的 fmap 编写为:

  instance (Functor f, Functor g) => Functor (Compose f g) where
    fmap f (Compose fga) = Compose $ fmap f (fmap f fga)

编译器提示:

* Couldn't match type `b' with `g b'
  `b' is a rigid type variable bound by
    the type signature for:
      fmap :: forall a b. (a -> b) -> Compose f g a -> Compose f g b
    at D:\haskell\chapter25\src\Twinplicative.hs:11:5
  Expected type: f (g b)
    Actual type: f b
* In the second argument of `($)', namely `fmap f (fmap f fga)'
  In the expression: Compose $ fmap f (fmap f fga)
  In an equation for `fmap':
      fmap f (Compose fga) = Compose $ fmap f (fmap f fga)
* Relevant bindings include
    fga :: f (g a)
      (bound at D:\haskell\chapter25\src\Twinplicative.hs:11:21)
    f :: a -> b
      (bound at D:\haskell\chapter25\src\Twinplicative.hs:11:10)
    fmap :: (a -> b) -> Compose f g a -> Compose f g b
      (bound at D:\haskell\chapter25\src\Twinplicative.hs:11:5)

如何在没有组合运算符的情况下组合上面的fmap？

Answer 1

您提供给 fmap 最左侧应用程序的函数应具有类型 g a -> g b，而您提供的 f 类型为 a -> b。您可以使用 fmap 即 fmap f 将函数 a -> b 提升为 g a -> g b。外部 fmap 的第二个参数应具有类型 f (g a)，即 fga 的类型:

instance (Functor f, Functor g) => Functor (Compose f g) where
    fmap f (Compose fga) = Compose $ fmap (fmap f) fga