[Haskell-cafe] Deconstruction

[Daniel Peebles]

You can't. The type can't be known, unfortunately.

With a wrapper like that you typically turn on rank-2 polymorphism and apply
a function to the value directly:

withBar :: Bar -> (forall a. BarLike a => a -> r) ->r
withBar (Bar x) = f x

Hope this helps,
Dan

On Sat, Dec 26, 2009 at 9:58 AM, haskell at kudling.de <haskell at kudling.de>wrote:

>  Hi,
>
>  while this works:
>
>
>   data Foo a = Foo a
>
>  unwrapFoo :: Foo a -> a
>  unwrapFoo (Foo x) = x
>
>
>  this:
>
>
>  {-# LANGUAGE ExistentialQuantification #-}
>
>   class BarLike a where
>      doSomething :: a -> Double
>
>   data Bar = forall a. BarLike a => Bar a
>
>  unwrapBar :: Bar -> a
>  unwrapBar (Bar x) = x
>
>
>  gives me:
>
>
>       Couldn't match expected type `a' against inferred type `a1'
>        `a' is a rigid type variable bound by
>            the type signature for `unwrapBar' at test.hs:8:20
>        `a1' is a rigid type variable bound by
>             the constructor `Bar' at test.hs:9:11
>      In the expression: x
>      In the definition of `unwrapBar': unwrapBar (Bar x) = x
>
>
>  How can i deconstruct the enclosed value of type a?
>
>  Thanks,
>  Lenny
>
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