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  • haskell network io hgetline

    - by Alex
    I want to read all the data on a handle, and then block waiting for more data. listen1 stops when there is a '\n' character in the stream. listen2 works and could be made completely general by imitating the code for hGetNonBlocking. What is the best way to do this? import qualified Data.ByteString as B loop = sequence_ . repeat listen1 :: Handle -> TChan B.ByteString -> IO() listen1 sock chan = do loop ( do s <- B.hGetLine sock atomically (writeTChan chan s) ) listen2 :: Handle -> TChan B.ByteString -> IO() listen2 sock chan = do loop ( do s <- B.hGet sock 1 s1 <- B.hGetNonBlocking sock 65000 atomically (writeTChan chan (B.append s s1)) )

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  • Haskell: Left-biased/short-circuiting function

    - by user2967411
    Two classes ago, our professor presented to us a Parser module. Here is the code: module Parser (Parser,parser,runParser,satisfy,char,string,many,many1,(+++)) where import Data.Char import Control.Monad import Control.Monad.State type Parser = StateT String [] runParser :: Parser a -> String -> [(a,String)] runParser = runStateT parser :: (String -> [(a,String)]) -> Parser a parser = StateT satisfy :: (Char -> Bool) -> Parser Char satisfy f = parser $ \s -> case s of [] -> [] a:as -> [(a,as) | f a] char :: Char -> Parser Char char = satisfy . (==) alpha,digit :: Parser Char alpha = satisfy isAlpha digit = satisfy isDigit string :: String -> Parser String string = mapM char infixr 5 +++ (+++) :: Parser a -> Parser a -> Parser a (+++) = mplus many, many1 :: Parser a -> Parser [a] many p = return [] +++ many1 p many1 p = liftM2 (:) p (many p) Today he gave us an assignment to introduce "a left-biased, or short-circuiting version of (+++)", called (<++). His hint was for us to consider the original implementation of (+++). When he first introduced +++ to us, this was the code he wrote, which I am going to call the original implementation: infixr 5 +++ (+++) :: Parser a -> Parser a -> Parser a p +++ q = Parser $ \s -> runParser p s ++ runParser q s I have been having tons of trouble since we were introduced to parsing and so it continues. I have tried/am considering two approaches. 1) Use the "original" implementation, as in p +++ q = Parser $ \s - runParser p s ++ runParser q s 2) Use the final implementation, as in (+++) = mplus Here are my questions: 1) The module will not compile if I use the original implementation. The error: Not in scope: data constructor 'Parser'. It compiles fine using (+++) = mplus. What is wrong with using the original implementation that is avoided by using the final implementation? 2) How do I check if the first Parser returns anything? Is something like (not (isNothing (Parser $ \s - runParser p s) on the right track? It seems like it should be easy but I have no idea. 3) Once I figure out how to check if the first Parser returns anything, if I am to base my code on the final implementation, would it be as easy as this?: -- if p returns something then p <++ q = mplus (Parser $ \s -> runParser p s) mzero -- else (<++) = mplus Best, Jeff

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  • Haskell Parsec Numeration

    - by Martin
    I'm using Text.ParserCombinators.Parsec and Text.XHtml to parse an input like this: - First type A\n -- First type B\n - Second type A\n -- First type B\n --Second type B\n And my output should be: <h11 First type A\n</h1 <h21.1 First type B\n</h2 <h12 Second type A\n</h2 <h22.1 First type B\n</h2 <h22.2 Second type B\n</h2 I have come to this part, but I cannot get any further: title1= do{ ;(count 1 (char '-')) ;s <- many1 anyChar newline ;return (h1 << s) } title2= do{ ;(count 2 (char '--')) ;s <- many1 anyChar newline ;return (h1 << s) } text=do { ;many (choice [try(title1),try(title2)]) } main :: IO () main = do t putStr "Error: " print err Right x - putStrLn $ prettyHtml x This is ok, but it does not include the numbering. Any ideas? Thanks!

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  • Custom whiteSpace using Haskell Parsec

    - by fryguybob
    I would like to use Parsec's makeTokenParser to build my parser, but I want to use my own definition of whiteSpace. Doing the following replaces whiteSpace with my definition, but all the lexeme parsers still use the old definition (e.g. P.identifier lexer will use the old whiteSpace). ... lexer :: P.TokenParser () lexer = l { P.whiteSpace = myWhiteSpace } where l = P.makeTokenParser myLanguageDef ... Looking at the code for makeTokenParser I think I understand why it works this way. I want to know if there are any workarounds to avoid completely duplicating the code for makeTokenParser?

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  • Haskell Parse Paragraph and em element with Parsec

    - by Martin
    I'm using Text.ParserCombinators.Parsec and Text.XHtml to parse an input like this: this is the beginning of the paragraph --this is an emphasized text-- and this is the end\n And my output should be: <p>this is the beginning of the paragraph <em>this is an emphasized text</em> and this is the end\n</p> This code parses and returns an emphasized element em = do{ ;count 2 (char '-') ; ;s <- manyTill anyChar (count 2 (char '-')) ;return (emphasize << s) } But I don't know how to get the paragraphs with emphasized items Any ideas? Thanks!!

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  • Multiple Haskell cabal-packages in one directory

    - by aleator
    What is the recommended way of having several cabal packages in one directory? Why: I have an old project with many separable modules. Since originally they formed just one program it was, and still is, handy to have them in same directory for easy compiling. Options Just suffer and split everything, including VCS holding the stuff, into different directories? Hack cabal until it is happy with multiple .cabal files in same directory? Make another subdirectory for each module and put .cabal files there along with symlinks to original pieces of code? Something smarter? What?

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  • Level-order in Haskell

    - by brain_damage
    I have a structure for a tree and I want to print the tree by levels. data Tree a = Nd a [Tree a] deriving Show type Nd = String tree = Nd "a" [Nd "b" [Nd "c" [], Nd "g" [Nd "h" [], Nd "i" [], Nd "j" [], Nd "k" []]], Nd "d" [Nd "f" []], Nd "e" [Nd "l" [Nd "n" [Nd "o" []]], Nd "m" []]] preorder (Nd x ts) = x : concatMap preorder ts postorder (Nd x ts) = (concatMap postorder ts) ++ [x] But how to do it by levels? "levels tree" should print ["a", "bde", "cgflm", "hijkn", "o"]. I think that "iterate" would be suitable function for the purpose, but I cannot come up with a solution how to use it. Would you help me, please?

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  • Wrong IO actions order using putStr and getLine

    - by QWRp
    I have a code : main = do putStr "Test input : " content <- getLine putStrLn content And when I run it (with runhaskell) or compile it (ghc 6.10.4) result is like this: asd Test input : asd I'm new to haskell and in my opinion printing should be first. Am I right? In code sample on http://learnyouahaskell.com/ which used putStr then getLine presented output is different than mine (IMHO correct). When I use putStrLn program works as expected (print then prompt and print). Is it a bug in ghc, or it is the way that it should work?

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  • [Haskell] Problem when mixing type classes and type families

    - by Giuseppe Maggiore
    Hi! This code compiles fine: {-# LANGUAGE MultiParamTypeClasses, FunctionalDependencies, FlexibleInstances, UndecidableInstances, FlexibleContexts, EmptyDataDecls, ScopedTypeVariables, TypeOperators, TypeSynonymInstances, TypeFamilies #-} class Sel a s b where type Res a s b :: * instance Sel a s b where type Res a s b = (s -> (b,s)) instance Sel a s (b->(c,a)) where type Res a s (b->(c,a)) = (b -> s -> (c,s)) but as soon as I add the R predicate ghc fails: {-# LANGUAGE MultiParamTypeClasses, FunctionalDependencies, FlexibleInstances, UndecidableInstances, FlexibleContexts, EmptyDataDecls, ScopedTypeVariables, TypeOperators, TypeSynonymInstances, TypeFamilies #-} class Sel a s b where type Res a s b :: * instance Sel a s b where type Res a s b = (s -> (b,s)) class R a where type Rec a :: * cons :: a -> Rec a elim :: Rec a -> a instance Sel a s (b->(c,Rec a)) where type Res a s (b->(c,Rec a)) = (b -> s -> (c,s)) complaining that: Illegal type synonym family application in instance: b -> (c, Rec a) In the instance declaration for `Sel a s (b -> (c, Rec a))' what does it mean and (most importantly) how do I fix it? Thanks

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  • How to convert ByteString to [Word8] in Haskell?

    - by bodacydo
    I am dealing with ByteStrings and at this one place I need to use [Word8] but I have a ByteString. How do I convert a ByteString to [Word8] list? I tried unpack function from ByteString library but it returned a [Char] list rather than [Word8] list. Do I need to take this step and convert it first to [Char] list and only to [Word8] list? If so, how do I do that. Please advise the most efficient method! Thank you!

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  • simple text menu in Haskell

    - by snorlaks
    Hello, I would like to know what is the best solution to create simple menu with functionality described below (pseudo code) just like im used to: while (true){ x = readLine(); case (x): x == "1" then do sth1 function x == "2" then do sth2 function } thanks for help, maybe any other ideas on how to make menu not in pattern described above?

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  • "Subclassing" show in Haskell?

    - by me2
    Lets say I have the following: data Greek = Alpha | Beta | Gamma | Phi deriving Show I want to use the default showing of all items except Beta, which I want to say "two". Can I do this?

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  • Haskell IO russian symbols

    - by Anton
    Hi. I try process file which writen by russian symbols. When read and after write text to file i get something like: "\160\192\231\229\240\225\224\233\228\230\224\237" How i can get normal symbols ? Thanks

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  • Updating List Elements, Haskell

    - by Tom
    I have homework where I am to update a list using a function that takes two elements and returns a value of part of the first element given in the function. So it's required to update the entire listing by going through each element and update its value by applying the function against all other elements in the list (including itself). So far I've been trying to firstly map the list (so that each element is done the same) and then specifically update each elements value by mapping again just the value of the specified element however in trying to map just the specific value through: the function, the specific element and the entire list I keep getting complaints that I'm inferring the list of values made from the 'map function p@list list' rather than simply giving the value at p@list. Is this the correct method to try to update a list against the entire list itself? EDIT: spelling mistakes and grammar

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  • Haskell parsec parsing a string of items

    - by Chris
    I have a list that I need to parse where the all but the last element needs to be parsed by one parser, and the last element needs to be parsed by another parser. a = "p1 p1b ... p2" or a = "p2" Originally I tried parser = do parse1 <- many parser1 parse2 <- parser2 return AParse parse1 parse2 The problem is that parse1 can consume a parse2 input. So parse1 always consumes the entire list, and leave parse2 with nothing. Is there a way to say to apply parse1 to everything besides the last element in a string, and then apply parse2?

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  • Haskell Input Return Tuple

    - by peterwkc
    Hello to all, i wonder can a IO() function return tuple because i would like to get these out of this function as input for another function. investinput :: IO()->([Char], Int) investinput = do putStrLn "Enter Username : " username <- getLine putStrLn "Enter Invest Amount : " tempamount <- getLIne let amount = show tempamount return (username, amount) Please help. Thanks.

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  • Haskell compile time function calculation

    - by egon
    I would like to precalculate values for a function at compile-time. Example (real function is more complex, didn't try compiling): base = 10 mymodulus n = n `mod` base -- or substitute with a function that takes -- too much to compute at runtime printmodules 0 = [mymodulus 0] printmodules z = (mymodulus z):(printmodules (z-1)) main = printmodules 64 I know that mymodulus n will be called only with n < 64 and I would like to precalculate mymodulus for n values of 0..64 at compile time. The reason is that mymodulus would be really expensive and will be reused multiple times.

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  • (type theoretical) How is ([] ==) [] typed in haskell?

    - by Ingo
    It sounds silly, but I can't get it. Why can the expression [] == [] be typed at all? More specifically, which type (in class Eq) is inferred to the type of list elements? In a ghci session, I see the following: Prelude> :t (==[]) (==[]) :: (Eq [a]) => [a] -> Bool But the constraint Eq [a] implies Eq a also, as is shown here: Prelude> (==[]) ([]::[IO ()]) <interactive>:1:1: No instance for (Eq (IO ())) arising from use of `==' at <interactive>:1:1-2 Probable fix: add an instance declaration for (Eq (IO ())) In the definition of `it': it = (== []) ([] :: [IO ()]) Thus, in []==[], the type checker must assume that the list element is some type a that is in class Eq. But which one? The type of [] is just [a], and this is certainly more general than Eq a = [a].

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  • Haskell optimization of a function looking for a bytestring terminator

    - by me2
    Profiling of some code showed that about 65% of the time I was inside the following code. What it does is use the Data.Binary.Get monad to walk through a bytestring looking for the terminator. If it detects 0xff, it checks if the next byte is 0x00. If it is, it drops the 0x00 and continues. If it is not 0x00, then it drops both bytes and the resulting list of bytes is converted to a bytestring and returned. Any obvious ways to optimize this? I can't see it. parseECS = f [] False where f acc ff = do b <- getWord8 if ff then if b == 0x00 then f (0xff:acc) False else return $ L.pack (reverse acc) else if b == 0xff then f acc True else f (b:acc) False

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  • Parsec Haskell Lists

    - by Martin
    I'm using Text.ParserCombinators.Parsec and Text.XHtml to parse an input and get a HTML output. If my input is: * First item, First level ** First item, Second level ** Second item, Second level * Second item, First level My output should be: <ul><li>First item, First level <ul><li>First item, Second level </li><li>Second item, Second level </li></ul></li><li>Second item, First level</li></ul> I wrote this, but obviously does not work recursively list= do{ s <- many1 item;return (olist << s) } item= do{ (count 1 (char '*')) ;s <- manyTill anyChar newline ;return ( li << s) } Any ideas? the recursion can be more than two levels Thanks!

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  • Help with possible Haskell type inference quiz questions

    - by Linda Cohen
    foldr:: (a -> b -> b) -> b -> [a] -> b map :: (a -> b) -> [a] -> [b] mys :: a -> a (.) :: (a -> b) -> (c -> a) -> c -> b what is inferred type of: a.map mys :: b.mys map :: c.foldr map :: d.foldr map.mys :: I've tried to create mys myself using mys n = n + 2 but the type of that is mys :: Num a => a -> a What's the difference between Num a = a - a and just a - a? Or what does 'Num a =' mean? Is it that mys would only take Num type? So anyways, a) I got [a] - [a] I think because it would just take a list and return it +2'd according to my definition of mys b) (a - b) - [a] - [b] I think because map still needs take both arguments like (*3) and a list then returns [a] which goes to mys and returns [b] c) I'm not sure how to do this 1. d) I'm not sure how to do this 1 either but map.mys means do mys first then map right? Are my answers and thoughts correct? If not, why not? THANKS!

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  • Parsec Haskell to HTML

    - by Martin
    I'm using Text.ParserCombinators.Parsec and Text.XHtml to parse an input like this: hello 123 --this is an emphasized text-- bye\n And my output should be: <p>hello 123 <em>this is an emphasized text</em> bye\n</p> Any ideas? Thanks!!

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