You can also export the type without exporting the constructors. That way "import"ers can use the type in type signatures and instance declarations while still not being able to use anything but the exported interface. E.g. instead of Module Set ( emptySet , makeSet , firstOfSet ) where use Module Set ( Set -- <=========== , emptySet , makeSet , firstOfSet ) where . (The non-hiding export is Module Set ( Set(..) -- <=========== , emptySet , makeSet , firstOfSet ) where .) This is covered in the Haskell report; see: http://www.haskell.org/onlinereport/modules.html#abstract-types . mike

At 13:15 2002-01-22 -0500, Hongwei Xi wrote:

...

<...> In Haskell, I guess that the one implemented later is always chosen. Why can't I have two different implementations for an interface?

Actually, I can't think of situations where I would desire this. Could you please give an example?

...

Another problem with Haskell classes is that there is currently no way of hiding type information. For instance, suppose that I want to implement a module for operations on sets but I do not want to reveal what data representation I use for sets. Is there a way of doing this in Haskell?

--Hongwei

There is: Use restricted exports. Only export functions on the datatype, and don't export the datatype itself. For example:

----------------------- Module Set ( emptySet , makeSet , firstOfSet ) where

-- Not exported, only locally visible data Set a = EmptySet | OneElementSet a

-- Using these functions, everyone can -- make, change and read Set's emptySet :: Set a emptySet = EmptySet

makeSet :: a -> Set a makeSet x = OneElementSet x

firstOfSet :: Set a -> a firstOfSet EmptySet = error "Set.firstOfSet: Empty set" firstOfSet (OneElementSet x) = x -----------------------

Regards,

Rijk-Jan van Haaften

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