问题描述
对于一个monad M，是否可以将A =>M[B] 变成 M[A =>B]?
For a monad M, Is it possible to turn A => M[B] into M[A => B]?
我试过遵循这些类型无济于事，这让我认为这是不可能的，但我想我还是会问.另外，在 Hoogle 中搜索 a ->m b ->m (a -> b) 没有返回任何东西，所以我运气不佳.
I've tried following the types to no avail, which makes me think it's not possible, but I thought I'd ask anyway. Also, searching Hoogle for a -> m b -> m (a -> b) didn't return anything, so I'm not holding out much luck.
推荐答案
实践中
不，它不能完成，至少不能以有意义的方式完成.
In Practice
No, it can not be done, at least not in a meaningful way.
考虑这个 Haskell 代码
Consider this Haskell code
action :: Int -> IO String
action n = print n >> getLine
这首先需要 n，打印它(这里执行 IO)，然后从用户那里读取一行.
This takes n first, prints it (IO performed here), then reads a line from the user.
假设我们有一个假设的 transform :: (a -> IO b) ->IO (a -> b).然后作为一个心理实验，考虑:
Assume we had an hypothetical transform :: (a -> IO b) -> IO (a -> b). Then as a mental experiment, consider:
action' :: IO (Int -> String)
action' = transform action
上面要提前做好所有的IO，才知道n，然后返回一个纯函数.这不能等价于上面的代码.
The above has to do all the IO in advance, before knowing n, and then return a pure function. This can not be equivalent to the code above.
为了强调这一点，请考虑下面这段无意义的代码:
To stress the point, consider this nonsense code below:
test :: IO ()
test = do f <- action'
putStr "enter n"
n <- readLn
putStrLn (f n)
神奇的是，action' 应该提前知道用户接下来要输入什么！一个会话看起来像
Magically, action' should know in advance what the user is going to type next! A session would look as
42 (printed by action')
hello (typed by the user when getLine runs)
enter n
42 (typed by the user when readLn runs)
hello (printed by test)
这需要时间机器，所以做不到.
This requires a time machine, so it can not be done.
不，这是不可能的.这个论点类似于我对类似问题提出的论点.
No, it can not be done. The argument is similar to the one I gave to a similar question.
矛盾假设 transform :: forall m a b.单子 m =>(a -> m b) ->m (a -> b) 存在.将 m 特化为 continuation monad ((_ -> r) -> r)(我省略了 newtype 包装器).
Assume by contradiction transform :: forall m a b. Monad m => (a -> m b) -> m (a -> b) exists.Specialize m to the continuation monad ((_ -> r) -> r) (I omit the newtype wrapper).
transform :: forall a b r. (a -> (b -> r) -> r) -> ((a -> b) -> r) -> r
专门化r=a:
transform :: forall a b. (a -> (b -> a) -> a) -> ((a -> b) -> a) -> a
申请:
transform const :: forall a b. ((a -> b) -> a) -> a
根据 Curry-Howard 同构，以下是一个直觉重言式
By the Curry-Howard isomorphism, the following is an intuitionistic tautology
((A -> B) -> A) -> A
但这是皮尔斯定律，在直觉逻辑中无法证明.矛盾.
but this is Peirce's Law, which is not provable in intuitionistic logic. Contradiction.
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