Re: Re : escape from existential quantification
> It occasionally happens that I *know* what type is (or at least ought to be) inside > an existential type, but unfortunately GHC doesn't, and I need to get the value out. > This can be solved using dynamic types, for example you declare the datatype as You can also use an unsafe cast operation if you know statically what the type will be, but the compiler doesn't. See for example http://www.isi.edu/~hdaume/haskell/NewBinary/TestBits.hs for an example of this. It essentially has a function which writes items of arbitrary type (in a list) to a binary memory location and then reads the back. When reading them back, the only way to know the type is to look at the type of the corresponding item on the "write" list. It relies on this to make the casting really a safe operation. ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re : escape from existential quantification
Wang Meng wrote I understand that existentially bound types cannot escape. For example, say we have data Foo = forall a. Foo Int a Then we cannot define a function extract (Foo i a) = a However,this limitation makes it extremly difficult to program with local quantifications.Is there any way to by pass this? It occasionally happens that I *know* what type is (or at least ought to be) inside an existential type, but unfortunately GHC doesn't, and I need to get the value out. This can be solved using dynamic types, for example you declare the datatype as > data Foo = forall a. Typeable a => Foo Int a (or something like that), then you can extract the a value by going to Dynamic and back. ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: escape from existential quantification
On Thu, 27 Feb 2003 18:26:31 + Keith Wansbrough <[EMAIL PROTECTED]> wrote: > > The idea is to use a type more like this: > > data Foo = forall a. Foo Int a (a -> (Int,Bool)) (a -> Int) (a -> > Foo) > > where the functions are the operations you want to use on the data Or else one can use type classes: - data Foo = forall a. Show a => Foo a instance Show Foo where show (Foo x) = show x main = print [Foo 3,Foo "ciao"] - Vincenzo -- Mai pensato a cosa vuol dire 10 anni di embargo? Si può chiedere ad una popolazione ormai allo stremo di subire un'altra guerra e magari un secondo embargo? Cosa c'entrano i bambini che muoiono di fame, di radiazioni e di mancanza di medicine con i giochi di potere di un dittatore? Che colpa ne portano? Leggere per esempio: http://www.fulviopoglio.com/salvi1.htm ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
Re: escape from existential quantification
> I understand that existentially bound types cannot escape. > > For example, say we have > data Foo = forall a. Foo Int a > > Then we cannot define a function > extract (Foo i a) = a > > However,this limitation makes it extremly difficult to program with local > quantifications.Is there any way to by pass this? The idea is to use a type more like this: data Foo = forall a. Foo Int a (a -> (Int,Bool)) (a -> Int) (a -> Foo) where the functions are the operations you want to use on the data. So now a list of Foos can contain data of many different types, as long as it is paired with the appropriate accessor functions for those types. You can use it like this: case x of Foo n x f g h -> if snd (f x) then g x else 0 for example. --KW 8-) -- Keith Wansbrough <[EMAIL PROTECTED]> http://www.cl.cam.ac.uk/users/kw217/ University of Cambridge Computer Laboratory. ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
escape from existential quantification
I understand that existentially bound types cannot escape. For example, say we have data Foo = forall a. Foo Int a Then we cannot define a function extract (Foo i a) = a However,this limitation makes it extremly difficult to program with local quantifications.Is there any way to by pass this? -W-M- @ @ | \_/ ___ Haskell mailing list [EMAIL PROTECTED] http://www.haskell.org/mailman/listinfo/haskell
