- Beginners - Haskell.org

Parsec 3.0 problems
by Ian Duncan 21 Apr '09

21 Apr '09

Hello everyone, I have the simple applicative parser combinator: parseLabel = Label <$> (between spaces (char ':') (many1 alphaNum)) This gives me the error: No instance for (Stream s m Char) arising from a use of `spaces' at tree.hs:18:32-37 Possible fix: add an instance declaration for (Stream s m Char) In the first argument of `between', namely `spaces' In the second argument of `(<$>)', namely `(between spaces (char ':') (many1 alphaNum))' In the expression: Label <$> (between spaces (char ':') (many1 alphaNum)) When I test this function in ghci, I have no problems, yet whenever I try to open it in a file it fails. What am I doing wrong? For those who are curious, Label is a data constructor in the following data type: data Loc = Label String | Addr Int deriving Show Thanks! Ian Duncan

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Question on Lists
by Nathan Holden 19 Apr '09

19 Apr '09

I have been reading Chris Okasaki's PhD thesis on Purely Functional Data Structures (www.cs.cmu.edu/~rwh/theses/*okasaki*.pdf) and it discusses his idea of lazy lists (He uses Standard ML in the paper), and it raised some questions for me. To cut to the chase, my question is this: Say I have list x of length n, and I have a single piece of data y. Does x ++ [y] take more cycles, or the same as if I'd said y:[x] (which would get it on the wrong side?)

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Installing Yi With GTK Frontend
by aditya siram 19 Apr '09

19 Apr '09

Hi all, I am having issues installing Yi with the GTK frontend. I have GTK2hs installed correctly but when I do 'yi --help' I get: Usage: yi [option...] [file] --self-check run self-checks -f [frontend] --frontend=[frontend] Select frontend, which can be one of: vty -y path --config-file=path Specify a configuration file -V --version Show version information -h --help Show this help --debug Write debug information in a log file -l [num] --line=[num] Start on line number --as=[editor] Start with editor keymap, where editor is one of: emacs, vim, cua --recompile-force Force recompile of custom yi before starting --resume Resume execution of yi from previous state --recompile Recompile custom yi if required then exit --ghc-options=[flags] Flags to pass to GHC Does Yi need a specific version of Gtk2hs? Thanks ... -deech

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Re: [Haskell-beginners] Partial Loading and Debugging with GHCI
by Kalman Noel 18 Apr '09

18 Apr '09

aditya siram schrieb: > I would then have to comment out 'bad', reload the program and query output > types to my hearts content. This is fine for a small program, but when I > have functions are downriver from 'bad', it gets extremely cumbersome. Is > there a way to load a file such that functions that compile stay loaded in > the interpreter even if something else fails? You can identify the offensive functions and write, in your example, bad :: String -- bad = hello ++ number bad = undefined

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Re: [Haskell-beginners] Installing Yi With GTK Frontend
by Kalman Noel 17 Apr '09

17 Apr '09

aditya siram schrieb: > I am having issues installing Yi with the GTK frontend. The gtk frontend has been replaced by the pango frontend. It depends on gtk2hs likewise. On installation, pass -fpango to Setup.hs to make sure that the frontend will get installed, or to be told what it doesn't find. Kalman

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Partial Loading and Debugging with GHCI
by aditya siram 17 Apr '09

17 Apr '09

Hi all, I'm curious as to how one would go about debugging the following program: hello :: String hello = "Hello" world :: String world = "World" number :: Int number = 1 helloWorld :: String helloWorld = hello ++ " " ++ world -- This fails! bad :: String bad = hello ++ number Predictably once I load this program into ghci I get the following error: [1 of 1] Compiling Main ( TestCode.hs, interpreted ) TestCode.hs:14:15: Couldn't match expected type `[Char]' against inferred type `Int' In the second argument of `(++)', namely `number' In the expression: hello ++ number In the definition of `bad': bad = hello ++ number Failed, modules loaded: none. At this point what I want to do is to query the output type of the different functions like so: > :t hello but GHCI won't let me because the program hasn't been loaded. I would then have to comment out 'bad', reload the program and query output types to my hearts content. This is fine for a small program, but when I have functions are downriver from 'bad', it gets extremely cumbersome. Is there a way to load a file such that functions that compile stay loaded in the interpreter even if something else fails? Thanks ... -deech

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Re: Thinking about monads (Brent Yorgey)
by Arthur Chan 16 Apr '09

16 Apr '09

You know, I was wondering... if Monads are a subset of Functors, and Applicative is a subset of Functors, and Monads are a subset of Applicative... shouldn't it be possible to tack on the definitions that automatically derive Functor and Applicative? Isn't it the case that there is really only one way to define Applicative for a Monad anyway? And isn't there only one way to define fmap for a Monad that makes sense? On Tue, Apr 14, 2009 at 2:39 PM, <beginners-request(a)haskell.org> wrote: > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://www.haskell.org/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Re: Re: Thinking about monads (Brent Yorgey) > 2. Re: How would you run a monad within another monad? > (Arthur Chan) > 3. Re: How would you run a monad within another monad? > (Arthur Chan) > 4. Re: How would you run a monad within another monad? > (Arthur Chan) > 5. Re: How would you run a monad within another monad? > (Jason Dusek) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Tue, 14 Apr 2009 16:42:08 -0400 > From: Brent Yorgey <byorgey(a)seas.upenn.edu> > Subject: Re: [Haskell-beginners] Re: Thinking about monads > To: beginners(a)haskell.org > Message-ID: <20090414204207.GA16671(a)seas.upenn.edu> > Content-Type: text/plain; charset=us-ascii > > On Mon, Apr 13, 2009 at 03:11:42PM -0700, Michael Mossey wrote: > > > > I know Maybe is both a functor and a monad, and I was thinking: what's > the > > difference? They are both wrappers on types. Then I realized, the > > difference is: they have different class definitions. > > In fact, every monad should be a functor, but not every functor is a > monad. Being a monad is a much stronger condition than being a functor. > > > > > class Functor f where > > fmap :: (a->b) -> f a -> f b > > > > (Note how fussy this definition would be in C++. It would be a kind of > > template, but would probably look a lot more complex and would require > > lengthy declarations.) > > > > class Monad m where > > a >>= b :: m a -> (a -> m b) -> m b > > Don't forget return :: a -> m a ! That's the other key method in the > Monad > class. (There are also >> and 'fail' but those are unimportant---the > first is just a specialization of >>=, and fail is a hack). > > -Brent > > > ------------------------------ > > Message: 2 > Date: Tue, 14 Apr 2009 14:05:22 -0700 > From: Arthur Chan <baguasquirrel(a)gmail.com> > Subject: Re: [Haskell-beginners] How would you run a monad within > another monad? > To: Jason Dusek <jason.dusek(a)gmail.com> > Cc: beginners(a)haskell.org > Message-ID: > <74cabd9e0904141405i1fbadb85u8b87ffb05d61c493(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > Here's my contrived example that threw the error. > > If you go into ghci, and do a `:t (foo' "blah" myDoohickey)`, you will get > the type signature "IO ()". > Doing the same for myOtherDoohickey returns "IO True" > > So you would think that you'd be able to uncomment the code that makes IO > an > instance of Toplevel. foo' is a function that allows IO to run monadic > values of type Doohickey. But it doesn't work. > > > --- > > import IO > import Control.Monad.Reader > > > class (Monad n) => Doohickey n where > putRecord :: String -> n () > > class (Monad m) => Toplevel m where > foo :: (Doohickey n) => FilePath -> n a -> m a > > newtype IOToplevelT a = IOToplevelT { runIOToplevelT :: ReaderT Handle IO a > } deriving (Monad, MonadReader Handle, MonadIO) > > instance Doohickey IOToplevelT where > putRecord = liftIO . putStrLn > > foo' s k = do > f <- liftIO $ openFile s AppendMode > runReaderT (runIOToplevelT k) f > > --instance Toplevel IO where > -- foo = foo' > > myDoohickey = do > putRecord "foo" > putRecord "bar" > > myOtherDoohickey = do > putRecord "hello" > putRecord "world" > return True > > > On Mon, Apr 13, 2009 at 7:55 PM, Jason Dusek <jason.dusek(a)gmail.com> > wrote: > > > Copypasting and loading your code doesn't throw an error. Please, > > pastebin an example that demonstrates the error. > > > > -- > > Jason Dusek > > >

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RWH Ch. 27 TCP Server Example
by aditya siram 15 Apr '09

15 Apr '09

Hi all, I am trying to understand the TCP Syslog Server example [1] from RWH. Specifically my question is on this piece of code: -- | Process incoming messages procMessages :: MVar () -> Socket -> SockAddr -> IO () procMessages lock connsock clientaddr = do connhdl <- socketToHandle connsock ReadMode hSetBuffering connhdl LineBuffering messages <- hGetContents connhdl mapM_ (handle lock clientaddr) (lines messages) hClose connhdl handle lock clientaddr "syslogtcpserver.hs: client disconnected" How does control stay on "mapM_ (handle lock clientaddr) (lines messages)" ? It would seem that the server would handle one message and immediately close the handle and end - but it doesn't. I'm guessing this has something to do with laziness, but I don't see how to apply it. thanks .. -deech [1] http://book.realworldhaskell.org/read/sockets-and-syslog.html

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How would you run a monad within another monad?
by Arthur Chan 14 Apr '09

14 Apr '09

Hey All, Suppose we have this generalized situation: I have some class that does IO with user data, but we want to hide the full IO interface, so we do something like this: class (Monad m) => Doohickey m where getCurrentTime :: m Time olderRecords :: m [String] addRecord :: String -> m () Suppose next that we want to run a monadic value of that type, with some context, inside another monad. For example, we have some file on disk with user data. We want the top level to open the file, and then run a monadic action with the above signature. The username provides the context. Give us a username, and we'll open a file, and then run a monadic value w/ the above signature on the file. e.g. class (Monad m) => TopLevel m where openFileAndRun :: (Doohickey n) => UserName -> n a -> m a In this example, we could have a function that takes an input string s, goes through the old records, and if it sees that s is not present, then it will add it to the records. Suppose then that we want to generalize the interface, so that it could apply equally to a local file, a db, the cloud, whatever. That's why we need the Toplevel monad. We want to separate "what we want to do with the user data" from "how we will store and modify the userdata". Toplevel actions therefore needs to be able to execute Doohickey actions. How would you guys do this? I'm trying to write the signatures but ghc borks on "Ambiguous type variable in the constraints" error. It appears to be caused by the usage of the Doohickey typeclass in the signature of Toplevel, when I actually try to write a monadic value of type Toplevel, that calls a monadic value of type Doohickey. Much thanks for any help! -Arthur

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Thinking about monads
by Michael Mossey 14 Apr '09

14 Apr '09

I'm getting a better grasp on monads, I think. My original problem, I think, was that I was still thinking imperatively. So when I saw this: class Monad m where (>>=) :: m a -> (a -> m b) -> m b I didn't understand the "big deal". I thought, okay so you "do" something with the function (a -> m b) and you "arrive" at m b. Now I realize that chaining a sequence of Monads via that function, a -> m b, means that a is available to any function further down the line, because it is an argument to a series of nested functions. So, doSomething = thing >>= \x -> thing2 >>= \y -> return (x,y) >>= produces a series of nested functions in which all the arguments of earlier functions are available to later functions: x and y are available to "return" because they are arguments of functions further up in the chain. This resembles imperative code in which any variable, once set, is available further down. Any clarifications welcome. -Mike

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