Fwd: UNPACK Existential datatype

[Roman Cheplyaka]

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?