(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 <v.dijk....@gmail.com> wrote:
> 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
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