Send Beginners mailing list submissions to
[email protected]
To subscribe or unsubscribe via the World Wide Web, visit
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
or, via email, send a message with subject or body 'help' to
[email protected]
You can reach the person managing the list at
[email protected]
When replying, please edit your Subject line so it is more specific
than "Re: Contents of Beginners digest..."
Today's Topics:
1. Re: Using scanl for Fibonacci sequence (Apoorv Ingle)
2. Monadic functions definitions for free monadic DSL (Sumit Raja)
3. Re: Monadic functions definitions for free monadic DSL
(David McBride)
----------------------------------------------------------------------
Message: 1
Date: Wed, 12 Oct 2016 17:24:22 -0500
From: Apoorv Ingle <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Using scanl for Fibonacci sequence
Message-ID: <[email protected]>
Content-Type: text/plain; charset="utf-8"
Hi Boon,
I guess it is difficult to comprehend what scanl exactly does here.
If you see the definition of scanl from hackage
<http://hackage.haskell.org/package/base-4.9.0.0/docs/Prelude.html#v:scanl>
scanl :: (b -> a -> b) -> b -> [a] -> [b]
scanl is similar to foldl, but returns a list of successive reduced values from
the left:
scanl f z [x1, x2, ...] == [z, z `f` x1, (z `f` x1) `f` x2, …]
Think of calculating a running sum of the first 10 positive integers, it can be
easily computed using scanl
λ> scanl (+) 0 [1..10]
[0,1,3,6,10,15,21,28,36,45,55]
For fibonacci, you calculate the nth number by adding n-1th number and n-2th
number.
Lets try to unfold the definition of fibs
fibs = 1 : scanl (+) 1 fibs
now first element of fibs is by definition, ofcourse
1 (say x1)
second element of fibs will be (the first element generated by scanl)
1 (i.e. z — say x2)
Third element of fibs will be calculated as (the second element generated by
scanl)
z `f` x1 = 1 + 1 = 2 (say x3)
Fourth element will be (and also the 3rd element generated by scanl)
(z `f` x1) `f` x2 = (1 + 1) + 1 = 3 (say x4)
Fifth element will be
((z `f` x1) `f` x2) `f` x3 = ((1 + 1) + 1) + 2 = 5 (say x5)
and so on..
The elegance is of course achieved because of the lazy evaluation of the
infinite list
Hope this makes it some what clear.
Regards,
Apoorv
> On 11-Oct-2016, at 23:23, Lai Boon Hui <[email protected]> wrote:
>
> Hi All,
>
> i am not very sure how this can work
> fibs = 1 : scanl (+) 1 fibs
>
> Appreciate it if someone can guide me through by showing me a few steps of
> the function evaluation
>
> --
> Best Regards,
> Boon Hui
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
-------------- next part --------------
An HTML attachment was scrubbed...
URL:
<http://mail.haskell.org/pipermail/beginners/attachments/20161012/33a59cd5/attachment-0001.html>
------------------------------
Message: 2
Date: Thu, 13 Oct 2016 13:15:43 +1100
From: Sumit Raja <[email protected]>
To: [email protected]
Subject: [Haskell-beginners] Monadic functions definitions for free
monadic DSL
Message-ID:
<CAD4nrSc3pZ-K72GBt3=frduziwp0utgsmx9rhggaeysb9ew...@mail.gmail.com>
Content-Type: text/plain; charset=UTF-8
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
------------------------------
Message: 3
Date: Thu, 13 Oct 2016 08:09:59 -0400
From: David McBride <[email protected]>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <[email protected]>
Subject: Re: [Haskell-beginners] Monadic functions definitions for
free monadic DSL
Message-ID:
<can+tr43m+_wi9jcozb8ty7ozjx-wk0wf2psk5ngancv0krc...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <[email protected]> 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
> _______________________________________________
> Beginners mailing list
> [email protected]
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL:
<http://mail.haskell.org/pipermail/beginners/attachments/20161013/87816ab1/attachment-0001.html>
------------------------------
Subject: Digest Footer
_______________________________________________
Beginners mailing list
[email protected]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
End of Beginners Digest, Vol 100, Issue 10
******************************************
