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<p class="MsoNormal">I think this is a very reasonable suggestion. It would take some work to implement, but nothing fundamental.<o:p></o:p></p>
<p class="MsoNormal"><o:p> </o:p></p>
<p class="MsoNormal">Simon<o:p></o:p></p>
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<p class="MsoNormal"><b><span lang="EN-US" style="font-size:11.0pt;font-family:"Calibri",sans-serif">From:</span></b><span lang="EN-US" style="font-size:11.0pt;font-family:"Calibri",sans-serif"> Glasgow-haskell-users [mailto:glasgow-haskell-users-bounces@haskell.org]
<b>On Behalf Of </b>Nicholas Clarke<br>
<b>Sent:</b> 20 January 2015 13:08<br>
<b>To:</b> glasgow-haskell-users@haskell.org<br>
<b>Subject:</b> Fwd: UNPACK Existential datatype<o:p></o:p></span></p>
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I'd like to be able to use the UNPACK pragma on an existentially quantified datatype. So as in the below example:<o:p></o:p></p>
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<o:p> </o:p></p>
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{-# LANGUAGE ExistentialQuantification #-}<o:p></o:p></p>
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<o:p> </o:p></p>
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data Foo = forall a. Show a => Foo !a<o:p></o:p></p>
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instance Show Foo where<o:p></o:p></p>
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show (Foo a) = "Foo! " ++ show a<o:p></o:p></p>
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<o:p> </o:p></p>
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data Bar =<o:p></o:p></p>
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Bar {-# UNPACK #-} !Foo<o:p></o:p></p>
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deriving (Show)<o:p></o:p></p>
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<o:p> </o:p></p>
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main :: IO ()<o:p></o:p></p>
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main = do<o:p></o:p></p>
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let foo = Foo "Hello"<o:p></o:p></p>
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bar = Bar foo<o:p></o:p></p>
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print bar<o:p></o:p></p>
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<o:p> </o:p></p>
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I would expect the `Foo` constructor to be unpacked into Bar, as if I had written:<o:p></o:p></p>
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<o:p> </o:p></p>
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data Bar = forall a. Show a => Bar !a<o:p></o:p></p>
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<o:p> </o:p></p>
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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?<o:p></o:p></p>
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<o:p> </o:p></p>
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Cheers,<o:p></o:p></p>
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<o:p> </o:p></p>
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Nick<o:p></o:p></p>
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<o:p> </o:p></p>
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<o:p> </o:p></p>
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