For a monad M, Is it possible to turn A => M[B] into M[A => B]?
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.
No, it can not be done, at least not in a meaningful way.
Consider this Haskell code
action :: Int -> IO String
action n = print n >> getLine
This takes n first, prints it (IO performed here), then reads a line from the user.
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
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.
To stress the point, consider this nonsense code below:
test :: IO ()
test = do f <- action'
putStr "enter n"
n <- readLn
putStrLn (f n)
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)
This requires a time machine, so it can not be done.
No, it can not be done. The argument is similar to the one I gave to a similar question.
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
Specialize r=a:
transform :: forall a b. (a -> (b -> a) -> a) -> ((a -> b) -> a) -> a
Apply:
transform const :: forall a b. ((a -> b) -> a) -> a
By the Curry-Howard isomorphism, the following is an intuitionistic tautology
((A -> B) -> A) -> A
but this is Peirce's Law, which is not provable in intuitionistic logic. Contradiction.