Re: brain explosion in polymorphic state monad

10 Oct 2002

      I'm not sure why it's doing that, but you can see (and fix!) the same
problem in a simpler case:

data Foo a = forall b . Foo a b

foo (Foo a _) f = 
    let Foo _ b = f a
    in  Foo a b

This causes the same error.  Presumably this has to do with the compiler
worrying about escaping variables or something.  I'm not sure.  There's a
workaround, though, which I bet will work in your case.  First we define:

refoo (Foo a _) (Foo _ b) = Foo a b

Then we redefine the foo function using this:

foo x@(Foo a _) f = refoo x (f a)

and we have a semantically identical, but now acceptable, function.

HTH

 - Hal

--
Hal Daume III

 "Computer science is no more about computers    | hdaume@isi.edu
  than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume

On Thu, 10 Oct 2002, mathieu wrote:
...
Hello,
I am trying to define a polymorphic state monad using glasgow extensions and I got a brain explosion of ghc when i try to compile it.
Here is the code :
newtype StateT s m a = MkStateT (s -> m (a, s))
instance Monad m => Monad (StateT s m) where
  return x = MkStateT (\s -> return (x, s))
  MkStateT m1 >>= k =
    MkStateT
    (\s0 -> do (a, s1) <- m1 s0
               let MkStateT m2 = k a
               m2 s1 )
data Thread a = forall b . MkThread (StateT (Thread b) [] a)
instance Monad Thread where
  return = MkThread . return
  MkThread p >>= k = MkThread ( do x <- p
                                   let MkThread p' = k x 
                                   p' )
I got this error :
   My brain just exploded.
    I can't handle pattern bindings for existentially-quantified constructors.
    In the binding group
        MkThread p' = k x
    In the first argument of `MkThread', namely
        `(do
            x <- p
            let MkThread p' = k x
            p')'
    In the definition of `>>=':
        MkThread (do
                    x <- p
                    let MkThread p' = k x
                    p')
How can i define (>>=) for my thread monad ?
Thanks in advance for any piece of advice,
Mathieu
-- 
There are only 10 types of people in the world:
Those who understand binary and those who don't.
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