Re: [Haskell-cafe] multiline strings in haskell?
On Wed, 11 Jan 2006, Michael Vanier wrote: Is there any support for multi-line string literals in Haskell? I've done a web search and come up empty. I'm thinking of using Haskell to generate web pages and having multi-line strings would be very useful. Do you mean unlines [first line, second line, third line] ? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re[2]: [Haskell-cafe] I/O and utf8
Hello Einar, Wednesday, January 11, 2006, 6:14:44 PM, you wrote: EK Do you plan on supporting things like HTTP where the character set EK is only known in the middle of the parsing? yes, it is supported, see Examples/Encoding.hs in the http://freearc.narod.ru/Binary.tar.gz : h - openWithEncoding latin1 = openBinaryFile test ReadMode print = vGetLine h vSetEncoding h utf8 print = vGetLine h vSetEncoding h latin1 print = vGetLine h vClose h it's not optimized currently. if you will need more speed - yell me -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Shootout favoring imperative code
Hello Einar, Wednesday, January 11, 2006, 6:06:56 PM, you wrote: My version of the packed string library does have an hGetLine. Don Stewart was merging my version with his fps at some point, Don - any news on that? EK Getting a fast FastPackedString will solve the problems with many EK benchmarks. btw, JHC's version of FPS uses slightly less memory (i don't remember, 8 or 12 bytes per each string) and i think must be faster (because it uses ByteArray# instead of Addr#). so, the best variant is to add hGetLine to John's library set arr x yv (arr,x) =: yv looks better ;) EK and so forth. Usually imperative solutions have something like EK a[i] += b[i], which currently is quite tedious and ugly to EK translate to MArrays. Now it would become combineTo a i (+) b i. you are definitely a Hal Daume's client, look at http://www.isi.edu/~hdaume/STPP/ -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] multiline strings in haskell?
On Jan 12, 2006, at 1:34 AM, Henning Thielemann wrote: On Wed, 11 Jan 2006, Michael Vanier wrote: Is there any support for multi-line string literals in Haskell? I've done a web search and come up empty. I'm thinking of using Haskell to generate web pages and having multi-line strings would be very useful. Do you mean unlines [first line, second line, third line] The original poster probably meant something like this: let foo = This is a long string Which does not end until the matching end quote. Common Lisp has strings like this. Quite convenient whenever you're communicating using lots of strings of data. For example, when running as a cgi script. Thanks, Jason ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] multiline strings in haskell?
On Thu, 12 Jan 2006, Jason Dagit wrote: On Jan 12, 2006, at 1:34 AM, Henning Thielemann wrote: On Wed, 11 Jan 2006, Michael Vanier wrote: Is there any support for multi-line string literals in Haskell? I've done a web search and come up empty. I'm thinking of using Haskell to generate web pages and having multi-line strings would be very useful. Do you mean unlines [first line, second line, third line] The original poster probably meant something like this: let foo = This is a long string Which does not end until the matching end quote. I don't see the need for it, since unlines [ first line, second line, third line] works as well. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] multiline strings in haskell?
On 1/12/06, Henning Thielemann [EMAIL PROTECTED] wrote: On Thu, 12 Jan 2006, Jason Dagit wrote: On Jan 12, 2006, at 1:34 AM, Henning Thielemann wrote: On Wed, 11 Jan 2006, Michael Vanier wrote: Is there any support for multi-line string literals in Haskell? I've done a web search and come up empty. I'm thinking of using Haskell to generate web pages and having multi-line strings would be very useful. Do you mean unlines [first line, second line, third line] The original poster probably meant something like this: let foo = This is a long string Which does not end until the matching end quote. I don't see the need for it, since unlines [ first line, second line, third line] works as well. Nevertheless Haskell supports multiline strings (although it seems like a lot of people don't know about it). You escape it using \ and then another \ where the string starts again. str = multi\ \line Preludestr multiline /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Chameneos
--- Aaron Denney [EMAIL PROTECTED] wrote: Are we off-topic for this mailing-list? I'd just like to respond to this: Anyways, your shootout, your hard work, your rules, but having rulings on what's acceptable be easier to find would be nice. People here have made the effort to develop programs for the shootout - I appreciate /their/ hard work. __ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] What does the Haskell type system do with show (1+2)?
What does the Haskell type system do with expressions such as these . . . ? show 1 show (1+2) The type of the subexpressions 1 and 1+2 are ambiguous since they have type (Num a) = a. I'm under the assumption before 1+2 is evaluated, the 1 and 2 must be coerced into a concrete type such as Int, Integer, Double, etc, and before show 1 is evaluated, the 1 must be coerced into a concrete type. Is my assumption correct? If so, how does Haskell know into which type to coerce the subexpressions? If I try to write a new function, my_show, which converts an _expression_ into a string representation that includes type information, I run into errors with expressions like show 1 and show (1+2) because of the type ambiguity. class (Show a) = My_show a where my_show :: a - String instance My_show Int where my_show a = show a ++ :: Int instance My_show Integer where my_show a = show a ++ :: Integer I can avoid the errors if I change it to my_show (1::Int) or my_show ((1+2)::Int). I'm wondering what the difference is between, my_show and Haskell's built-in show that causes my_show to produce an error message when it is used with ambiguous types, but Haskell's show works okay with ambiguous types. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] What does the Haskell type system do with show (1+2)?
http://www.haskell.org/onlinereport/decls.html#default-decls http://www.haskell.org/tutorial/numbers.html#sect10.4 On 1/12/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: What does the Haskell type system do with expressions such as these . . . ? show 1 show (1+2) The type of the subexpressions 1 and 1+2 are ambiguous since they have type (Num a) = a. I'm under the assumption before 1+2 is evaluated, the 1 and 2 must be coerced into a concrete type such as Int, Integer, Double, etc, and before show 1 is evaluated, the 1 must be coerced into a concrete type. Is my assumption correct? If so, how does Haskell know into which type to coerce the subexpressions? If I try to write a new function, my_show, which converts an expression into a string representation that includes type information, I run into errors with expressions like show 1 and show (1+2) because of the type ambiguity. class (Show a) = My_show a where my_show :: a - String instance My_show Int where my_show a = show a ++ :: Int instance My_show Integer where my_show a = show a ++ :: Integer I can avoid the errors if I change it to my_show (1::Int) or my_show ((1+2)::Int). I'm wondering what the difference is between, my_show and Haskell's built-in show that causes my_show to produce an error message when it is used with ambiguous types, but Haskell's show works okay with ambiguous types. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- [EMAIL PROTECTED] http://www.updike.org/~jared/ reverse )-: ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Does anybody have a simple example of using continuation Monad?
I'm struggling with this example: http://www.nomaware.com/monads/html/contmonad.html#example After looking at it for the fourth time I got much more.. but still not enough.. because there are so much new things (when beeing translated into some kind of condition ? thentodo : elsetodo which is using ThenElse It wouldn't be any problem if the next example wasn't using continuation, too.. and that's about combining monads which is important, isn't it? At the tutorial there was mentioned that continuation monads are used for continuation passing style which I've looked up in wikipedia meaning something like splitting a task into different parts beeing executed delayed (for example because of user interaction filling a web form?) I think one simple example like ((+1).(\x-x**2)) in continuation style would make me understand a lot more.. Marc ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Does anybody have a simple example of using continuation Monad?
On 12/01/06, Marc Weber [EMAIL PROTECTED] wrote: I'm struggling with this example: http://www.nomaware.com/monads/html/contmonad.html#example After looking at it for the fourth time I got much more.. but still not enough.. because there are so much new things (when beeing translated into some kind of condition ? thentodo : elsetodo which is using ThenElse It wouldn't be any problem if the next example wasn't using continuation, too.. and that's about combining monads which is important, isn't it? At the tutorial there was mentioned that continuation monads are used for continuation passing style which I've looked up in wikipedia meaning something like splitting a task into different parts beeing executed delayed (for example because of user interaction filling a web form?) I think one simple example like ((+1).(\x-x**2)) in continuation style would make me understand a lot more.. Marc Yeah, that's probably what bothers me most about All About Monads, as it's otherwise quite a good tutorial. Continuations are a strange concept, and not exactly the first thing that beginners need to see. The continuation monad/transformer has its place, but it's rarely needed, and when abused, it just results in an unreadable mess. The basic idea about the continuation monad is that the entire computation you are defining is parametrised on a function which will take its result and continue to operate on it (the 'future'). The computation is built up in this way -- at each stage, we extend the computation by providing another piece of the future, while still leaving the computation as a whole parametrised on it. Normally we'd be forced to manage these futures by having explicit parameters for them and such, but the monad machinery hides all of this so that you don't have to worry so much about it. If you don't ever make use of the extra feature that you can get a handle on the future, you can use it just like the identity monad: addOne x = return (x+1) square x = return (x**2) f x = do y - addOne x; square y With explicit continuations, this would look something like: addOne x k = k (x + 1) square x k = k (x ** 2) f x k = addOne x (\v - square v k) Now, this seems like an awkward way to handle things as we're not making any use of the fact that at each stage, we have a handle to the future which can be used multiple times, passed into other functions, etc. The Cont monad gives one primitive for capturing the current continuation and passing it into a computation, called callCC. callCC :: ((a - Cont b) - Cont a) - Cont a This type requires some study to understand at first, but essentially, callCC takes a function from a future (a - Cont b) to a new computation (Cont a), and passes it the current future (which is accessible due to the funny way in which we're parametrising our computations). In terms of callCC, we can write other, more convenient ways to manipulate futures. The following is due to Tomasz Zielonka [1]: getCC :: MonadCont m = m (m a) getCC = callCC (\c - let x = c x in return x) getCC' :: MonadCont m = a - m (a, a - m b) getCC' x0 = callCC (\c - let f x = c (x, f) in return (x0, f)) getCC will get the current continuation explicitly as a computation which can be executed. This essentially gives us a 'goto-label' at that point in the computation, and executing it will jump back. This isn't terribly useful in plain Cont, except to land us in an infinite loop, but over a state monad, or IO, we can cause side-effect havoc, observe the state, and decide whether to return to the goto-label or not. Stealing an example from Tomasz' original message: -- prints hello! in an endless loop test :: IO () test = (`runContT` return) $ do jump - getCC lift $ putStrLn hello! jump getCC' is similar, but actually allows an additional parameter to be sent back. The parameter to getCC' is just the initial value. Here's a simplistic implementation of mod by repeated addition/subtraction which prints intermediate results as it goes, in the ContT transformed IO monad. x `modulo` m = (`runContT` return) $ do (u, jump) - getCC' x lift $ print u case u of _ | u 0 - jump (u + m) | u = m- jump (u - m) | otherwise - return u - Cale [1] http://www.haskell.org/pipermail/haskell-cafe/2005-July/010623.html ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
