Re: [Haskell-cafe] Resource module

On Wed, Jun 2, 2010 at 2:28 PM, Bas van Dijk wrote:

...

Before answering your questions I would like to make sure I understand your Resource type. When I want to create a memory Resource for example is the following what you have in mind?

{-# LANGUAGE Rank2Types #-}

-- from base: import Foreign.Ptr ( Ptr ) import Foreign.Marshal.Alloc ( mallocBytes, free )

-- from transformers: import Control.Monad.IO.Class ( liftIO )

-- from MonadCatchIO-transformers: import Control.Monad.CatchIO ( MonadCatchIO, bracket )

newtype Resource cap m = Resource { with :: forall a. (cap -> m a) -> m a }

type Memory m a = Resource (Ptr a) m

memory :: MonadCatchIO m => Int -> Memory m a memory size = Resource $ bracket (liftIO $ mallocBytes size) (liftIO . free)

Yes, exactly. I also create type aliases for resources providing a specific capability. On Wed, 2 Jun 2010 14:45:08 +0200, Bas van Dijk wrote:

The previous can also be generalized using my Resource class:

-- from regions: import qualified Control.Resource as C ( Resource(..) )

resource :: (MonadCatchIO m, C.Resource resource) => resource -> Resource (C.Handle resource) m resource r = Resource $ bracket (liftIO $ C.open r) (liftIO . C.close)

Yes, definitely. (This is not a literal generalisation of the 'memory' function, unless you make 'Int' an instance of 'C.Resource'; one would probably create a special type 'data Memory = Memory Int' instead. This is the difference I alluded to in my earlier email.) Regards, Arie