这是关于由Applicative和Monad组合两个简单算术运算符(两个偏函数(。到目前为止,我大致理解它。
-- ┌──────────────────────────────────────┐
-- | instance Applicative ((->) r) where |
-- | pure = const |
-- | (<*>) f g x = f x (g x) | (1)
-- | liftA2 q f g x = q (f x) (g x) |
-- | |
-- | instance Monad ((->) r) where |
-- | f >>= k = r -> k (f r) r | (2)
-- └──────────────────────────────────────┘
λ> (+) <*> (*2) $ 10
-- Type check:
(<*>) :: f (a -> b) -> f a -> f b
(-> r) (a -> b) -> (-> r) a -> (-> r) b
(r -> a -> b) -> (r -> a) -> (r -> b)
--------------- ---------
(+) ~ f (*2) ~ g r ~ x (10)
-- Actual calcuation: by (1)
f x (g x) = (+) 10 ((*2) 10) = (+) 10 20 = 30
λ> (*2) >>= (+) $ 10 -- 30
-- Type check:
(>>=) :: m a -> a -> m b -> m b
(-> r) a -> (-> r) a -> b -> (-> r) b
(r -> a) -> (a -> r -> b) -> (r -> b)
--------- ---------------
(*2) ~ f (+) ~ k r ~ 10
-- Actual calculation: by (2)
k (f r) r = (+) ((*2) 10) 10 = (+) 20 10 = 30
但当我试图将这些东西应用于某个结构时(也许(,我被卡住了
-- ┌────────────────────────────────────────────────┐
-- | instance Applicative Maybe where |
-- | pure = Just |
-- | Just f <*> m = fmap f m |
-- | Nothing <*> _m = Nothing |
-- | liftA2 f (Just x) (Just y) = Just (f x y) |
-- | liftA2 _ _ _ = Nothing |
-- | Just _m1 *> m2 = m2 |
-- | Nothing *> _m2 = Nothing |
-- | |
-- | instance Monad Maybe where |
-- | (Just x) >>= k = k x |
-- | Nothing >>= _ = Nothing |
-- | (>>) = (*>) |
-- └────────────────────────────────────────────────┘
λ> Just 10 >>= return . (+1) -- Just 11
-- Type check:
(>>=) :: m a -> a -> m b -> m b
---------- --------------
Just 10 return . (+1) :: a -> m b
so, m ~ Maybe, a ~ Int
Just 10 >>= return . (+1) :: m b
Maybe Int
-- Actual calculation:
Just 10 >>= return . (+1) = return . (+1) $ 10
= Just . (+1) $ 10
= Just 11
λ> Just >>= return (+1) $ 10 -- 11
-- This is point-free style
I don't get it.
-- Type check:
(>>=) :: m a -> a -> m b -> m b
---------- --------------
(-> a) (Maybe a) m (a -> a) <==== HERE!
I can't derive the correct type.
-- Actual calculation: <==== and HERE!
m >>= f = x -> f (m x) x <==== How can this formula be derived?
= (return (+1)) (Just 10) 10 <==== How can this be calculated?
莫纳德的表达有一种不带点的风格。我不明白。我怎么能像前面的简单算术表达式那样推导类型并得到结果呢?
非常感谢。
谢谢你的回答,尼尔。在你的帮助下,我可以找到我的思想和代码中的错误。但是我仍然不能正确地得到Just >>= return (+1)
的最终类型。我更新了我的问题,并试图推断出它的类型。我知道我的类型推断是错误的。我能得到更多的帮助来找到错误的部分并修复吗?
-- Type check: Incorrect
(>>=) :: m a -> a -> m b -> m b
(-> r) a -> (-> r) a -> b -> (-> r) b
-------- --------------
f k r -> k (f r) r
m f r -> m (f r) r
(-> d) (Maybe d) (-> r) (n -> n)
so, a ~ Maybe d
a ~ n, b ~ n -- This means a, b, n are all the same type.
-- Character a, b can be interchangeable.
Just >>= return (+1) :: (-> r) b
= (-> r) a -- by `a ~ b`
= (-> r) (Maybe d) -- by `a ~ Maybe d`
-- HERE: Is this right?
Should this be `(-> r) n`?
= a -> Maybe b -- by changing characters
HERE: WRONG RESULT TYPE??? It must be `a -> a` not `a -> Maybe a`
-- Actual calcuation:
Moand instance for function (`(-> r)`) type here (I got)
m >>= f = x -> f (m x) x
= return (+1) (Just 10) 10
= return (+1) (Just 10) $ 10
= const (+1) (Just 10) $ 10 -- by (1), pure = const
= (+1) $ 10 -- 'Just 10' was ignored by 'const (pure)' function.
= (+) 10 = 11
非常感谢。
在中
Just >>= return (+1) $ 10
Just
是函数r -> Maybe r
,因此使用函数monad((->) r)
,将其简化为
return (1+) (Just 10) 10
因为这是((->) r)
monad的>>=
的定义(正如您在文章顶部给出的f >>= k = r -> k (f r) r
(。
现在,return (1+)
应用于Just 10
,所以它也是一个函数,
return :: Monad m => a -> m a
(1+) :: Num m => n -> n
return (1+) :: (Monad m, Num n) => m (n -> n)
Num n => (r -> (n -> n))
Just 10 :: (Num i => Maybe i) ~ r
对于函数,return == const
,所以我们有
const (1+) (Just 10) 10
===
(1+) 10
===
11
你的另一个问题是,为什么((->) r)
monad,m >>= f = x -> f (m x) x
?(您确实注意到,它与顶部的定义f >>= k = r -> k (f r) r
相同,直到一些变量重命名,对吧?(
答案是,因为类型适合:
(>>=) :: m a -> (a -> m b ) -> m b
~ (r -> a) -> (a -> (r -> b)) -> (r -> b)
(>>=) m f x = b
where
mx = m x :: a
f_mx = f mx :: r -> b
b = f_mx x :: b
edit:让我们试着像编译器一样,不假思索地推导Just >>= return (+1)
的类型。我们将使用类型推导的主要规则,对应于逻辑中的Modus-Ponens规则:
A :: a
F :: t -> b
--------------------
F A :: b , t ~ a
诀窍是从一开始就使用所有不同类型的变量,这样我们甚至没有机会将它们混合在一起:
--Just>>=返回(+1(===(>>=(Just(返回(+1((return::Monad f=>d->f d(+1(::Num n=>n->n----------------return(+1(::(Monad f,Num n(=>f(n->n(---------------------------------------------------------(>>=(::莫纳德m=>m a->(a->m b(->m b只是:(->(c(也许c(----------------m~(->(c,a~也许c--------------------------(>>=(只是::莫纳德((->(c(=>(也许c->(->(cb(->cb~莫纳德((->(c(=>(可能是c->(c->b((->(c->c(----------------------return(+1(::(Monad f,Num n(=>f(n->n(--------------------f~(->((可能是c(,(n->n(~(c->b((>>=(仅(return(1+((:(莫纳德((->(c(,莫纳德(【->】(可能是c((=>c->b~Num n=>n->n