New type error when compiling with GHC head

[David Peixotto]

Ah, that worked perfectly! Thanks!

On Sep 16, 2010, at 11:45 AM, Simon Peyton-Jones wrote:

> For reasons explained in “Let should not be generalised” GHC isn’t going to generalise local let-bindings, at least not when  you switch on GADTs , which you have. 
>  
> So unBM is monomorphic, hence the error.
>  
> Give it a type signature, or make it top-level. Either works fine.
>  
> Simon
>  
> From: cvs-ghc-bounces at haskell.org [mailto:cvs-ghc-bounces at haskell.org] On Behalf Of David Peixotto
> Sent: 16 September 2010 17:38
> To: cvs-ghc at haskell.org
> Subject: New type error when compiling with GHC head
>  
> I'm now getting a type error when compiling with GHC HEAD that I wasn't getting before (also GHC 6.10 and GHC 6.12 work fine). The code is part of a larger program that I did not write and the type error is confusing to me. I cut it down to the minimal I could, so hopefully it will be useful. Can someone comment if this is a bug, or the program needs to be changed?
>  
> Code:(Backtrack.hs)
> {-# OPTIONS_GHC -fglasgow-exts #-}
> module Backtrack where
> import Control.Monad
>  
> -- Combining endomorphisms and continuations
> -- a la Ralf Hinze
>  
> -- BacktrackM = state monad transformer over the backtracking monad
> newtype BacktrackM s a = BM (forall b . (a -> s -> b -> b) -> s -> b -> b)
>  
> instance Monad (BacktrackM s) where
>     return a   = BM (\c s b -> c a s b)
>     BM m >>= k = BM (\c s b -> m (\a s b -> unBM (k a) c s b) s b)
>       where unBM (BM m) = m
>     fail _ = mzero
>  
> instance MonadPlus (BacktrackM s) where
>   mzero                 = BM (\c s b -> b)
>   (BM f) `mplus` (BM g) = BM (\c s b -> g c s $! f c s b)
>  
> Error:
> Backtrack.hs:13:18:
>     Couldn't match type `b' with `b2'
>       because this skolem type variable would escape: `b2'
>     This skolem is bound by
>       the polymorphic type `forall b. (a -> s -> b -> b) -> s -> b -> b'
>     The following variables have types that mention b
>       unBM :: BacktrackM s b1 -> (b1 -> s -> b -> b) -> s -> b -> b
>         (bound at Backtrack.hs:14:13)
>     In the first argument of `BM', namely
>       `(\ c s b -> m (\ a s b -> unBM (k a) c s b) s b)'
>     In the expression:
>       BM (\ c s b -> m (\ a s b -> unBM (k a) c s b) s b)
>     In an equation for `>>=':
>         (BM m) >>= k
>           = BM (\ c s b -> m (\ a s b -> unBM (k a) c s b) s b)
>           where
>               unBM (BM m) = m
>  
> -David
>  

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