[Haskell-beginners] Monadic functions definitions for free monadic DSL

David McBride toad3k at gmail.com
Thu Oct 13 12:09:59 UTC 2016


I would really like to help you, but without your imports, packages, etc,
it is really hard to interpret your program.  Like where does decodeUtf8
come from, or receive, or TCPSocket?  If they are functions you wrote, I
don't need their code, the types would be sufficient.

On Wed, Oct 12, 2016 at 10:15 PM, Sumit Raja <sumitraja at gmail.com> wrote:

> Hello,
>
> I am trying to get my head around free monads by developing a simple
> network abstraction DSL.
> I've made good progress before adding TCP/IP semantics of accepting
> connections. I'm now stuck with the creation of monadic functions.
>
> I've defined the following:
>
>     data NetworkActivity chan next = Accept chan next (chan -> next) |
>             Send chan ByteString (Bool -> next) |
>             Recv chan (ByteString -> next) |
>             Close chan (() -> next)
>
>     clse :: a -> Free (NetworkActivity a) Text
>     clse chan = liftF (Close chan (const "Quit"))
>
>     chatterServer :: a -> Free (NetworkActivity a) Text
>     chatterServer svrchan = Free $ Accept svrchan (chatterServer
> svrchan) chatterLoop
>
>     chatterLoop :: a -> Free (NetworkActivity a) Text
>     chatterLoop chan = Free $ Recv chan $ \bs -> case BS.uncons bs of
>       Nothing -> clse chan
>       Just x -> if bs == "Bye" then
>           Free $ Close chan (\_ -> Pure "Quit")
>         else
>           Free (Send chan bs (\_ -> chatterLoop chan))
>
> This works fine with the interpretTCP interpreter below accepting
> multiple connections:
>
>     interpretTCP :: Free (NetworkActivity TCPSocket) r -> IO r
>     interpretTCP prg = case prg of
>       Free (Accept serverSock svrLoop acceptProc) -> bracket (return
> serverSock)
>         (\s-> interpretTCP (clse s))
>         (\s-> do
>           (ss, sa) <- accept s
>           forkIO $ do
>             _ <- interpretTCP (acceptProc ss)
>             return ()
>           interpretTCP svrLoop
>         )
>       Free (Recv sock g) -> do
>         bs <- receive sock 4096 mempty
>         putStrLn (decodeUtf8 bs)
>         interpretTCP (g bs)
>       Free (Close sock g) -> do
>         close sock
>         putStrLn ("Server bye!" :: Text)
>         interpretTCP (g ())
>       Pure r -> return r
>       Free (Send sock pl g) -> do
>         sent <- send sock pl mempty
>         interpretTCP (g (sent > 0))
>
> Where I'm stuck is defining the monadic version of accept and I'm
> beginning to think my original
> data type defined above may be wrong. As an initial step I've defined
> the following:
>
>     recv :: a -> Free (NetworkActivity a) ByteString
>     recv chan = liftF (Recv chan identity)
>
>     sendit :: a -> ByteString -> Free (NetworkActivity a) Bool
>     sendit chan pl = liftF (Send chan pl identity)
>
>     mchatterServer :: a -> Free (NetworkActivity a) Text
>     mchatterServer chan = Free $ Accept chan (mchatterServer chan)
>                                                                    (\s
> -> return (identity s) >>= mchatterLoop)
>
> mchatterServer works as is, the interpreter accepts multiple
> connections. Similarly all good with recv and sendit.
> I am struggling with converting the Accept in mchatterServer into a
> function to use in the do syntax. The signature I think I should be
> using is
>
>     acc :: a -> NetworkActivity a Text -> Free (NetworkActivity a)
> (NetworkActivity a Text)
>
> What I can't figure out is why it can't follow the pattern of recv and
> sendit above:
>
>     acc chan next = liftF $ Accept chan next identity
>
> Which results in error on identity (using Protolude):
>
>     Expected type: a -> NetworkActivity a Text
>     Actual type: NetworkActivity a Text -> NetworkActivity a Text
>
> I can't really see how to get the types to line up and have now can't
> see through the type fog. What am I missing in my reasoning about the
> types?
>
> Help much appreciated!
>
> Thanks
>
> Sumit
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