Oops, I thought I had sent a response to the cafe, but it looks like it just went to Matthew. Unfortunately, I was trying to give a simplification of the real problem, where the monad is STM instead of []. Based on apfelmus's observation of why they can't be isomorphic, I'm guessing I'm out of luck. http://www.haskell.org/pipermail/haskell-cafe/2006-December/020041.html So in reality, I'm trying to construct something like f :: (a -> STM b) -> STM (a -> b) I just figured it was a general monadic kind of problem, more simply expressed using lists. But the (!!) solution doesn't make sense in this context. On 2/2/07, Yitzchak Gale wrote:

Chad Scherrer wrote:

...

Are (a -> [b]) and [a -> b] isomorphic? I'm trying to construct a function

f :: (a -> [b]) -> [a -> b]

that is the (at least one-sided) inverse of

f' :: [a -> b] -> a -> [b] f' gs x = map ($ x) gs

Anything better than this?

f g = [\x -> g x !! n | n <- [0..]]

-Yitz

-- Chad Scherrer "Time flies like an arrow; fruit flies like a banana" -- Groucho Marx

Uh, apologies. I got confused between reading your post and playing for a while, and answered the wrong question. /g On 2/2/07, J. Garrett Morris wrote:

On 2/2/07, Chad Scherrer wrote:

...

So in reality, I'm trying to construct something like f :: (a -> STM b) -> STM (a -> b)

I just figured it was a general monadic kind of problem, more simply expressed using lists. But the (!!) solution doesn't make sense in this context.

Perhaps this will work for you:

f x = join . liftM f

This typechecks, and seems to work as expected, e.g.:

Prelude Control.Concurrent.STM Control.Monad> atomically (f readTVar (newTVar 'a')) 'a'

/g

-- It is myself I have never met, whose face is pasted on the underside of my mind.

-- It is myself I have never met, whose face is pasted on the underside of my mind.

On Sat, Feb 03, 2007 at 10:13:17AM +0100, apfelmus@quantentunnel.de wrote:

Chad Scherrer wrote:

...

So in reality, I'm trying to construct something like f :: (a -> STM b) -> STM (a -> b)

Indeed, such an f most likely does not exist.

Yes, consider: f readTVar :: STM (TVar c -> c) That would be pretty dangerous. Best regards Tomasz