通用类型组合的可遍历实例
Traversable instance for general type composition
我完全坚持这是一本优秀的 Haskell Programming 书中的练习。
给定以下类型组合的新类型以及 Functor 和 Applicative 的实例,编写 Traversable (Compose f g)
的实例。
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
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
pure = Compose <$> pure . pure
Compose f <*> Compose x =
Compose $ ((<*>) <$> f) <*> x
我提出的解决方案看起来应该可行,基于 traverse.traverse
的类型,但 ghci 抱怨。我有一种模糊的感觉,这与 Compose
构造函数中的重新包装有关:
instance (Traversable f, Traversable g) => Traversable (Compose f g) where
traverse f1 (Compose fga) = (traverse.traverse) f1 fga
给出类型错误:
composing_types.hs:69:31:
Couldn't match type ‘b’ with ‘g b’
‘b’ is a rigid type variable bound by
the type signature for
traverse :: Applicative f1 =>
(a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
at composing_types.hs:69:3
Expected type: f1 (Compose f g b)
Actual type: f1 (Compose f g (g b))
Relevant bindings include
fga :: f (g a) (bound at composing_types.hs:69:24)
f1 :: a -> f1 b (bound at composing_types.hs:69:12)
traverse :: (a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
(bound at composing_types.hs:69:3)
In the expression: (traverse . traverse) f1 fga
In an equation for ‘traverse’:
traverse f1 (Compose fga) = (traverse . traverse) f1 fga
composing_types.hs:69:54:
Couldn't match type ‘f’ with ‘Compose f g’
‘f’ is a rigid type variable bound by
the instance declaration at composing_types.hs:68:10
Expected type: Compose f g (g a)
Actual type: f (g a)
Relevant bindings include
fga :: f (g a) (bound at composing_types.hs:69:24)
traverse :: (a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
(bound at composing_types.hs:69:3)
In the second argument of ‘traverse . traverse’, namely ‘fga’
In the expression: (traverse . traverse) f1 fga
H O L E S
这是另一个可以用漏洞表达式解决的好问题。
首先,假设我们已经定义了所有可折叠实例。
λ> instance (Foldable f, Foldable g) => Foldable (Compose f g) where
foldr = undefined
接下来,实例Traversable。 Compose
参数上的模式匹配,因为你知道你必须这样做,否则就把所有东西都留在洞里。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) = _ tua
GHC 会帮助吐出一个错误–
<interactive>:...:...
Found hole ‘_’ with type: f (Compose t u b)
–除了范围内所有变量的类型。
Relevant bindings include
tua :: t (u a) (bound at ...)
a2fb :: a -> f b (bound at ...)
traverse :: (a -> f b) -> Compose t u a -> f (Compose t u b)
(bound at ...)
(我已经选择了类型和值的名称,以便一切都整齐排列。不要注意幕后的那个人。)现在是时候的问题了:如何创造一个 [=20= 的值] 给定其他一切。我们知道
构造Compose t u b
的唯一方法是创建一个值t (u b)
。
除了 (1) pure
和 (2) fmap
之外,没有办法产生 f anything
的值,而且凭直觉我们知道我们不能使用 pure
因为我们试图在这里收集 a2fb :: a -> f b
的 'side effects'。
这引导我们找到下一个解决方案。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose (_ tua)
<interactive>:...
Found hole ‘_’ with type: t (u a) -> f (t (u b))
我们终于有了 t
。我们知道 t
是可遍历的,所以让我们尝试遍历它。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse _ tua) tua)
<interactive>:56:138:
Found hole ‘_’ with type: u a -> f (u b)
同样的交易。我们知道 u
是可遍历的,所以让我们尝试遍历它。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse (\ua -> traverse _ ua) tua) tua)
<interactive>:57:155:
Found hole ‘_’ with type: a -> f b
我们的 a2fb
.
的金凤花洞
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse (\ua -> traverse a2fb ua) tua) tua)
Eta-reduce 去除 lambda,最后得到 the solution。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose (traverse (traverse a2fb) tua)
我完全坚持这是一本优秀的 Haskell Programming 书中的练习。
给定以下类型组合的新类型以及 Functor 和 Applicative 的实例,编写 Traversable (Compose f g)
的实例。
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
instance (Applicative f, Applicative g) => Applicative (Compose f g) where
pure = Compose <$> pure . pure
Compose f <*> Compose x =
Compose $ ((<*>) <$> f) <*> x
我提出的解决方案看起来应该可行,基于 traverse.traverse
的类型,但 ghci 抱怨。我有一种模糊的感觉,这与 Compose
构造函数中的重新包装有关:
instance (Traversable f, Traversable g) => Traversable (Compose f g) where
traverse f1 (Compose fga) = (traverse.traverse) f1 fga
给出类型错误:
composing_types.hs:69:31:
Couldn't match type ‘b’ with ‘g b’
‘b’ is a rigid type variable bound by
the type signature for
traverse :: Applicative f1 =>
(a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
at composing_types.hs:69:3
Expected type: f1 (Compose f g b)
Actual type: f1 (Compose f g (g b))
Relevant bindings include
fga :: f (g a) (bound at composing_types.hs:69:24)
f1 :: a -> f1 b (bound at composing_types.hs:69:12)
traverse :: (a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
(bound at composing_types.hs:69:3)
In the expression: (traverse . traverse) f1 fga
In an equation for ‘traverse’:
traverse f1 (Compose fga) = (traverse . traverse) f1 fga
composing_types.hs:69:54:
Couldn't match type ‘f’ with ‘Compose f g’
‘f’ is a rigid type variable bound by
the instance declaration at composing_types.hs:68:10
Expected type: Compose f g (g a)
Actual type: f (g a)
Relevant bindings include
fga :: f (g a) (bound at composing_types.hs:69:24)
traverse :: (a -> f1 b) -> Compose f g a -> f1 (Compose f g b)
(bound at composing_types.hs:69:3)
In the second argument of ‘traverse . traverse’, namely ‘fga’
In the expression: (traverse . traverse) f1 fga
H O L E S
这是另一个可以用漏洞表达式解决的好问题。
首先,假设我们已经定义了所有可折叠实例。
λ> instance (Foldable f, Foldable g) => Foldable (Compose f g) where
foldr = undefined
接下来,实例Traversable。 Compose
参数上的模式匹配,因为你知道你必须这样做,否则就把所有东西都留在洞里。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) = _ tua
GHC 会帮助吐出一个错误–
<interactive>:...:...
Found hole ‘_’ with type: f (Compose t u b)
–除了范围内所有变量的类型。
Relevant bindings include
tua :: t (u a) (bound at ...)
a2fb :: a -> f b (bound at ...)
traverse :: (a -> f b) -> Compose t u a -> f (Compose t u b)
(bound at ...)
(我已经选择了类型和值的名称,以便一切都整齐排列。不要注意幕后的那个人。)现在是时候的问题了:如何创造一个 [=20= 的值] 给定其他一切。我们知道
构造
Compose t u b
的唯一方法是创建一个值t (u b)
。除了 (1)
pure
和 (2)fmap
之外,没有办法产生f anything
的值,而且凭直觉我们知道我们不能使用pure
因为我们试图在这里收集a2fb :: a -> f b
的 'side effects'。
这引导我们找到下一个解决方案。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose (_ tua)
<interactive>:...
Found hole ‘_’ with type: t (u a) -> f (t (u b))
我们终于有了 t
。我们知道 t
是可遍历的,所以让我们尝试遍历它。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse _ tua) tua)
<interactive>:56:138:
Found hole ‘_’ with type: u a -> f (u b)
同样的交易。我们知道 u
是可遍历的,所以让我们尝试遍历它。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse (\ua -> traverse _ ua) tua) tua)
<interactive>:57:155:
Found hole ‘_’ with type: a -> f b
我们的 a2fb
.
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose ((\tua -> traverse (\ua -> traverse a2fb ua) tua) tua)
Eta-reduce 去除 lambda,最后得到 the solution。
λ> instance (Traversable t, Traversable u) => Traversable (Compose t u) where
traverse a2fb (Compose tua) =
fmap Compose (traverse (traverse a2fb) tua)