Haskell 从应用角度实现幺半群
介绍以下练习: 根据单位和(**)实现纯and(),反之亦然 这里是Haskell 从应用角度实现幺半群,haskell,Haskell,介绍以下练习: 根据单位和(**)实现纯and(),反之亦然 这里是单项式和MyApplicative: class Functor f => Monoidal f where u :: f () -- using `u` rather than `unit` dotdot :: f a -> f b -> f (a,b) -- using instead of `(**)` class Functor f =
单项式
和MyApplicative
:
class Functor f => Monoidal f where
u :: f () -- using `u` rather than `unit`
dotdot :: f a -> f b -> f (a,b) -- using instead of `(**)`
class Functor f => MyApplicative f where
p :: a -> f a -- using instead of `pure`
apply :: f (a -> b) -> f a -> f b -- using instead of `(<**>)`
instance Monoidal Option where
u = p ()
dotdot None _ = None
dotdot _ None = None
dotdot (Some x) (Some y) = Some id <*> Some (x, y)
然后,我定义了实例MyApplicationOption
:
instance MyApplicative Option where
p = Some
apply None _ = None
apply _ None = None
apply (Some g) f = fmap g f
最后,我尝试在MyApplicative
的p
和apply
方面实现Monoidal选项
:
class Functor f => Monoidal f where
u :: f () -- using `u` rather than `unit`
dotdot :: f a -> f b -> f (a,b) -- using instead of `(**)`
class Functor f => MyApplicative f where
p :: a -> f a -- using instead of `pure`
apply :: f (a -> b) -> f a -> f b -- using instead of `(<**>)`
instance Monoidal Option where
u = p ()
dotdot None _ = None
dotdot _ None = None
dotdot (Some x) (Some y) = Some id <*> Some (x, y)
特别是,我很好奇如何用Applicative的
(
)正确地实现dotdot::fa->fb->f(a,b)
——在我的例子中,它是apply
Applicative
是单体
的一种简洁的替代表示形式。这两个类型类是等效的,您可以在这两个类型之间进行转换,而无需考虑特定的数据类型,如选项
。Applicative
的“简洁的备选演示文稿”基于以下两种等效形式
pure a = fmap (const a) unit
unit = pure ()
ff <*> fa = fmap (\(f,a) -> f a) $ ff ** fa
fa ** fb = pure (,) <*> fa <*> fb
替换为纯
,我们可以将类型()
和构造函数():()
替换为以恢复单元
pure :: a -> f a
pure () :: f ()
同样地(虽然不是那么简单)将类型(a,b)
和构造函数(,)::a->b->(a,b)
替换为liftA2
,以恢复**
liftA2 :: (a -> b -> c) -> f a -> f b -> f c
liftA2 (,) :: f a -> f b -> f (a,b)
Applicative
然后通过将函数application($)::(a->b)->a->b
提升到函子中来获取nice
操作符
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = liftA2 ($)
在下面答案的开头,我们很快解释了其中的大部分内容,解释得很好,谢谢。要不要在这里寄信用卡?
(<*>) :: f (a -> b) -> f a -> f b
(<*>) = liftA2 ($)
liftA2 :: Applicative f => (a -> b -> c) -> f a -> f b -> f c
liftA2 f a b = f <$> a <*> b