On Wed, 23 May 2007 19:54:27 +0100 "Neil Mitchell" <[EMAIL PROTECTED]> wrote: > Hi > > > > foo = do (1 :: Int) > > > > While intuitively this should be disallowed, it seems a pity that > > desugaring couldn't be totally separated from typechecking. Hmm. > > You can always desugar as: > > do x ==> return () >> x > > Although then you are relying on the Monad laws more than you possibly > should. You could also have: > > monady :: Monad m => m a -> m a > monady x = x > > do x ==> monady x
How about: do x ==> (x :: Monad m => m a) Or even: do x ==> (asTypeOf x (return ())) Cheers, Spencer Janssen _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
