Re: [Haskell-cafe] Control.Monad.Cont fun

[Andrew Pimlott]

On Thu, Jul 07, 2005 at 07:08:23PM +0200, Tomasz Zielonka wrote:
 Hello!
 
 Some time ago I wanted to return the escape continuation out of the
 callCC block, like this:
 
   getCC = callCC (\c - return c)
 
 But of course this wouldn't compile.
 
 I thought that it would be useful to be able to keep the current
 continuation and resume it later, like you can in do in scheme. Well,
 it was easy to do this in Haskell in the (ContT r IO) monad, by using an
 IORef. But I wasn't able to solve this for all MonadCont monads.
 
 After more then year of on-and-off trials, I've finally managed to do
 this! ;-)
 
   import Control.Monad.Cont
 
   getCC :: MonadCont m = m (m a)
   getCC = callCC (\c - let x = c x in return x)

I was inspired by this message, and thought I could simply this further
as

getCC = callCC (\c - let x = c x in x)

After all, c can give us a MonadCont with any result type.  The first
problem is that callCC is not polymorphic enough, but that is easily
fixed[1] (concerning myself just with Cont):

ccc :: ((a - (forall b. Cont r b)) - Cont r a) - Cont r a
ccc f = Cont (\k - runCont (f (\x - Cont (\_ - k x))) k)

ccc is just callCC with a more polymorphic type.  But still, try as I
might, I could not get getCC to typecheck, no matter what type
annotations I used.  This works:

getCC' :: Cont r (Cont r a)
getCC' = ccc (\c - let x = c x in c x)

But eg

getCC :: Cont r (Cont r a)
getCC = ccc (\(c :: Cont r a - (forall b. Cont r b)) -
let x :: forall b. Cont r b = c x in x)

gives

Couldn't match `forall b. Cont r a - Cont r b'
   against `forall b. Cont r a - Cont r b'
In a lambda abstraction:
\ (c :: Cont r a - (forall b. Cont r b))
- let x :: forall b. Cont r b = c x in x
In the first argument of `ccc', namely
`(\ (c :: Cont r a - (forall b. Cont r b))
  - let x :: forall b. Cont r b = ... in x)'
In the definition of `getCC'':
getCC' = ccc (\ (c :: Cont r a - (forall b. Cont r b))
  - let x :: forall b. Cont r b = ... in x)

for which I have no riposte.  Is this a bug?  Is there any way to type
this expression?  (I am using ghc 6.4.)

Andrew

[1] http://www.haskell.org/hawiki/ContinuationsDoneRight
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Re: [Haskell-cafe] Control.Monad.Cont fun

[Thomas Jäger]

Hello Andrew,

On 7/25/05, Andrew Pimlott [EMAIL PROTECTED] wrote:
 getCC :: Cont r (Cont r a)
 getCC = ccc (\(c :: Cont r a - (forall b. Cont r b)) -
 let x :: forall b. Cont r b = c x in x)

 gives

 [Error]
ghc-6.2 accepts this:
  getCC :: Cont r (Cont r a)
  getCC = ccc (\(c :: Cont r a - (forall b. Cont r b)) -
   let x :: forall b. Cont r b; x = c x in c x)

ghc-6.4/6.5 will also give the mysterious error above, but the
following works fine, thanks to scoped type variables:
  getCC :: forall a r. Cont r (Cont r a)
  getCC = ccc (\c -
   let x :: forall b. Cont r b; x = c x in x)

 for which I have no riposte.  Is this a bug?
I would think so. I can't find anything in the documentation that
disallows such polymorphic type annotations.

Thomas
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