Re: [Haskell-cafe] Why this existential type could not work?

Magicloud Magiclouds schrieb am Thu Aug 21 2014 at 08:37:09:

Hi,

For example, code like this:

newtype Sealed = Sealed { unSealed :: forall a. SomeClass a => TVar a }

I think this should be written as data Sealed = forall a. SomeClass a => Sealed { unSealed :: TVar a }