Move MonadIO to base

[Tyson Whitehead]

On April 23, 2010 21:55:52 David Menendez wrote:
> N.B. The instance given for ContT does not do what you want. Here's
> the definition again:
>
>    morphIO f = ContT $ \k -> morphIO (\d -> f (\m -> d (runContT m k)))
>
> Note that the entire continuation is passed to d. If you are defining,
> say, block in terms of morphIO
>
>    gblock m = morphIO (\d -> block (d m))
>
> Then m and its continuation get masked, not just m. Similar problems
> occur for backtracking monads and iteratees.

I think you have to redo ContT so the underlying monad is exposed to the 
continuation instead of being permanently trapped inside a closure.

newtype ContT m a = ContT { runContT :: forall r. (m a -> m r) -> m r }

class MorphIO m where
    morphIO :: (forall b. (m a -> IO b) -> IO b) -> m a

instance Monad m => Monad (ContT m) where
    return x = ContT $ \k -> k $ return x
    m >>= f  = ContT $ \k -> runContT m (>>= \x -> runContT (f x) k)

instance MorphIO m => MorphIO (ContT m) where
    morphIO f = ContT $ \k -> k $ morphIO $ \d -> f $ \m -> d $ runContT m id

Now just the single desired computation is being passed to d, which is then 
re-wrapped and passed onto the continuation.

Cheers!  -Tyson
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