
How is parsing of the *NOUNPACK* pragma relevant here? On 23/01/15 10:45, Alexander V Vershilov wrote:
Hi.
As far as I understand it was fixed as:
commit 1d32a8503c2ebfab2bbdb696fe65dd0823d1ed27 Author: Simon Peyton Jones
Date: Mon Dec 1 17:07:48 2014 +0000 Fix parser for UNPACK pragmas
{-# NOUNPACK #-} {-# NOUNPACK #-} ! were being parsed the same way. The former was wrong.
Thanks to Alan Zimmerman for pointing this out
So it will fix is in 7.10. And I can't reproduce this anymore on ghc-HEAD.
On 20 January 2015 at 17:35, Roman Cheplyaka
wrote: Interesting question. I managed to trace this to:
compiler/basicTypes/MkId.hs:699
isUnpackableType fam_envs ty | Just (tc, _) <- splitTyConApp_maybe ty , Just con <- tyConSingleAlgDataCon_maybe tc , isVanillaDataCon con = ok_con_args (unitNameSet (getName tc)) con | otherwise = False
where isVanillaDataCon is defined as:
dcVanilla :: Bool, -- True <=> This is a vanilla Haskell 98 data constructor -- Its type is of form -- forall a1..an . t1 -> ... tm -> T a1..an -- No existentials, no coercions, nothing.
There's no explanation why this limitation is introduced; it might be just a conservative one.
On 20/01/15 15:08, Nicholas Clarke wrote:
I'd like to be able to use the UNPACK pragma on an existentially quantified datatype. So as in the below example:
{-# LANGUAGE ExistentialQuantification #-}
data Foo = forall a. Show a => Foo !a instance Show Foo where show (Foo a) = "Foo! " ++ show a
data Bar = Bar {-# UNPACK #-} !Foo deriving (Show)
main :: IO () main = do let foo = Foo "Hello" bar = Bar foo print bar
I would expect the `Foo` constructor to be unpacked into Bar, as if I had written:
data Bar = forall a. Show a => Bar !a
However, instead I get the 'Ignoring unusable UNPACK pragma on the first argument of ‘Bar’' warning. Is there a reason this shouldn't work, or a workaround to get it to do so?
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