Hello Tomasz,
This stuff is very interesting! At first sight, your definition of
getCC seems quite odd, but it can in fact be derived from its
implementation in an untyped language.
On 7/7/05, Tomasz Zielonka
Some time ago I wanted to return the escape continuation out of the callCC block, like this:
getCC = callCC (\c -> return c) We get the error message
test124.hs:8:29: Occurs check: cannot construct the infinite type: t = t -> t1 Expected type: t -> t1 Inferred type: (t -> t1) -> m b Haskell doesn't support infinite types, but we can get close enough by creating a type C m b such that C m b and C m b -> m b become isomorphic: newtype C m b = C { runC :: C m b -> m b } With the help of C we can implement another version of getCC and rewrite the original example. getCC1 :: (MonadCont m) => m (C m b) getCC1 = callCC $ \k -> return (C k) test1 :: ContT r IO () test1 = do jmp <- getCC1 liftIO $ putStrLn "hello1" jmp `runC` jmp -- throw the continuation itself, -- so we can jump to the same point the next time. return () We can move the self-application of jmp into getCC to get the same type signature as your solution, but we still rely on the auxiliary datatype C. getCC2 :: MonadCont m => m (m b) getCC2 = do jmp <- callCC $ \k -> return (C k) return $ jmp `runC` jmp In order to move the function (\jmp -> jmp `runC` jmp) into callCC, the following law, that all instances of MonadCont seem to satisfy, is very helpful. f =<< callCC g === callCC (\k -> f =<< g ((=<<) k . f)) In particular (regarding Functor as superclass of Monad), it follows f `fmap` callCC g === callCC (\k -> f `fmap` g (k . f)) Therefore, getCC2 is equivalent to getCC3 :: MonadCont m => m (m b) getCC3 = callCC $ \k -> return (selfApp $ C (k . selfApp)) where selfApp :: C m b -> m b selfApp jmp = jmp `runC` jmp It is easy to get rid of C here arriving exactly at your definition of getCC.
getCC :: MonadCont m => m (m a) getCC = callCC (\c -> let x = c x in return x)
getCC' :: MonadCont m => a -> m (a, a -> m b) getCC' x0 = callCC (\c -> let f x = c (x, f) in return (x0, f)) For what it's worth, this can be derived in much the same way from the (not well-typed)
getCC' x = callCC $ \k -> return (k, x) using the auxillary type newtype C' m a b = C' { runC' :: (C' m a b, a) -> m b }
Besides sharing my happiness, I want to ask some questions:
- is it possible to define a MonadFix instance for Cont / ContT?
It must be possible to define something that looks like a MonadFix instance, after all you can define generally recursive functions in languages like scheme and sml which "live in a ContT r IO monad", but this has all kinds of nasty consequences, iirc. Levent Erkök's thesis suggests (pp. 66) that there's no implementation of mfix that satisfies the purity law. http://www.cse.ogi.edu/PacSoft/projects/rmb/erkok-thesis.pdf
- do you think it would be a good idea to add them to Control.Monad.Cont? I think so, because they simplify the use of continuations in an imperative setting and are probably helpful in understanding continuations. Letting continuations escape is quite a common pattern in scheme code, and painful to do in Haskell without your cool trick. I'd also like to have shift and reset functions there :)
Best wishes, Thomas