(resending this to the list because this failed yesterday because of
the mailinglist downtime)
On Sat, Oct 30, 2010 at 1:57 AM, Bas van Dijk
I could isolate it a bit more if you want.
And so I did. The following is another instance of the problem I'm having but set in a more familiar setting: {-# LANGUAGE RankNTypes #-} import Control.Monad.ST foo :: (forall s. ST s a) -> a foo st = ($) runST st Couldn't match expected type `forall s. ST s a' with actual type `ST s a' In the second argument of `($)', namely `st' Note that: 'foo st = runST st' type checks as expected. Surprisingly 'foo st = runST $ st' also type checks! I find the latter surprising because according to the report[1]: e1 op e2 = (op) e1 e2. So either both should type check or both should fail. I guess that a RULE somewhere eliminates the ($) before the type-checker kicks in. I do find that a little strange because I thought RULES where applied after type checking. Regards, Bas [1] http://www.haskell.org/onlinereport/exps.html#operators