Fwd: UNPACK Existential datatype

[Roman Cheplyaka]

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 <simonpj at microsoft.com>
> 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 <roma at ro-che.info> 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?
>>
>> _______________________________________________
>> Glasgow-haskell-users mailing list
>> Glasgow-haskell-users at haskell.org
>> http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
> 
> 
>