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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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  • 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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  • 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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  • 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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  • NILL value in haskell

    - by snorlaks
    Hello, I get input (x) from user, convert it to Int by let y = (read x)::Int and then I would like the function to behave in a special way if user gave nothing (empty string). -- In this place I would like to handle situation in which user -- gave empty string as argument -- this doesnt work :/ yearFilter [] y = True --This works fine as far as y is integer yearFilter x y | x == (objectYear y) = True | otherwise = False Thanks for help, Bye

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  • Very slow guards in my monadic random implementation (haskell)

    - by danpriduha
    Hi! I was tried to write one random number generator implementation, based on number class. I also add there Monad and MonadPlus instance. What mean "MonadPlus" and why I add this instance? Because of I want to use guards like here: -- test.hs -- import RandomMonad import Control.Monad import System.Random x = Rand (randomR (1 ::Integer, 3)) ::Rand StdGen Integer y = do a <-x guard (a /=2) guard (a /=1) return a here comes RandomMonad.hs file contents: -- RandomMonad.hs -- module RandomMonad where import Control.Monad import System.Random import Data.List data RandomGen g => Rand g a = Rand (g ->(a,g)) | RandZero instance (Show g, RandomGen g) => Monad (Rand g) where return x = Rand (\g ->(x,g)) (RandZero)>>= _ = RandZero (Rand argTransformer)>>=(parametricRandom) = Rand funTransformer where funTransformer g | isZero x = funTransformer g1 | otherwise = (getRandom x g1,getGen x g1) where x = parametricRandom val (val,g1) = argTransformer g isZero RandZero = True isZero _ = False instance (Show g, RandomGen g) => MonadPlus (Rand g) where mzero = RandZero RandZero `mplus` x = x x `mplus` RandZero = x x `mplus` y = x getRandom :: RandomGen g => Rand g a ->g ->a getRandom (Rand f) g = (fst (f g)) getGen :: RandomGen g => Rand g a ->g -> g getGen (Rand f) g = snd (f g) when I run ghci interpreter, and give following command getRandom y (mkStdGen 2000000000) I can see memory overflow on my computer (1G). It's not expected, and if I delete one guard, it works very fast. Why in this case it works too slow? What I do wrong?

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  • Understanding this matrix transposition function in Haskell

    - by dmindreader
    This matrix transposition function works, but I'm trying to understand its step by step execurtion and I don't get it. transpose:: [[a]]->[[a]] transpose ([]:_) = [] transpose x = (map head x) : transpose (map tail x) with transpose [[1,2,3],[4,5,6],[7,8,9]] it returns: [[1,4,7],[2,5,8],[3,6,9]] I don't get how the concatenation operator is working with map. It is concatenating each head of x in the same function call? How?

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  • installation HDBC-SQlite3 Haskell

    - by snorlaks
    Hello, Im facing the problem during installation : setup configure Configuring HDBC-sqlite3-2.3.0.0... setup: Missing dependency on a foreign library: * Missing C library: sqlite3 This problem can usually be solved by installing the system package that provides this library (you may need the "-dev" version). If the library is already installed but in a non-standard location then you can use the flags --extra-include-dirs= and --extra-lib-dirs= to specify where it is. what should I do ? thanks for any help

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  • Better data stream reading in Haskell

    - by Tim Perry
    I am trying to parse an input stream where the first line tells me how many lines of data there are. I'm ending up with the following code, and it works, but I think there is a better way. Is there? main = do numCases <- getLine proc $ read numCases proc :: Integer -> IO () proc numCases | numCases == 0 = return () | otherwise = do str <- getLine putStrLn $ findNextPalin str proc (numCases - 1) Note: The code solves the Sphere problem https://www.spoj.pl/problems/PALIN/ but I didn't think posting the rest of the code would impact the discussion of what to do here.

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  • Haskell Ord instance with a Set

    - by mvid
    I have some code that I would like to use to append an edge to a Node data structure: import Data.Set (Set) import qualified Data.Set as Set data Node = Vertex String (Set Node) deriving Show addEdge :: Node -> Node -> Node addEdge (Vertex name neighbors) destination | Set.null neighbors = Vertex name (Set.singleton destination) | otherwise = Vertex name (Set.insert destination neighbors) However when I try to compile I get this error: No instance for (Ord Node) arising from a use of `Set.insert' As far as I can tell, Set.insert expects nothing but a value and a set to insert it into. What is this Ord?

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  • haskell recursive function

    - by snorlaks
    Hello, Please help me with writing function which takes two arguments : list of ints and index (int) and returns list of integers with negative of value on specified index position in the table MyReverse :: [Int]-Int-[Int] for example myReverse [1,2,3,4,5] 3 = [1,2,-3,4,5] if index is bigger then length of the list or smaller then 0 return the same list. Thanks for help

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  • Haskell Type error

    - by Jon
    I am getting a Couldn't match expected type error on this code and am not sure why. Would appreciate it if someone could point me in the right direction as to fixing it. import qualified Data.ByteString.Lazy as S import Data.Binary.Get import Data.Word getBinary :: Get Word16 getBinary = do a <- getWord16be "Test.class" return (a) main :: IO () main = do contents <- S.getContents print getBinary contents Specifically it cannot match expected type 'S.ByteString - IO ()' to inferred type 'IO ()'

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  • Haskell - Parsec Parsing <p> element

    - by Martin
    I'm using Text.ParserCombinators.Parsec and Text.XHtml to parse an input like this: This is the first paragraph example\n with two lines\n \n And this is the second paragraph\n And my output should be: <p>This is the first paragraph example\n with two lines\n</p> <p>And this is the second paragraph\n</p> I defined: line= do{ ;t<-manyTill (anyChar) newline ;return t } paragraph = do{ t<-many1 (line) ;return ( p << t ) } But it returns: <p>This is the first paragraph example\n with two lines\n\n And this is the second paragraph\n</p> What is wrong? Any ideas? Thanks!

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  • Problem understanding treesort in Haskell

    - by Jerry
    I am trying to figure out how exactly does treesort from here work (I understand flatten, insert and foldr). I suppose what's being done in treesort is applying insert for each element on the list thus generating a tree and then flattening it. The only problem I can't overcome here is where the list (that is the argument of the function) is hiding (because it is not written anywhere as an argument except for the function type declaration). One more thing: since dot operator is function composition, why is it an error when I change: treesort = flatten . foldr insert Leaf to treesort = flatten( foldr insert Leaf )?

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  • Haskell - function (that returns a list) on each element in a list

    - by Ben
    The assignment is to create a multiples function and I essentially want todo the following code: map (\t -> scanl (\x y -> x+y) t (repeat t)) listofnumbers The problem is that the scanl function returns a list of results rather than the one which the map function requires. So is there a function that will allow the return of lists?

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  • Haskell: variant of `show` that doesn't wrap String and Char in quotes

    - by Joey Adams
    I'd like a variant of show (let's call it label) that acts just like show, except that it doesn't wrap Strings in " " or Chars in ' '. Examples: > label 5 "5" > label "hello" "hello" > label 'c' "c" I tried implementing this manually, but I ran into some walls. Here is what I tried: {-# LANGUAGE FlexibleInstances #-} {-# LANGUAGE UndecidableInstances #-} module Label where class (Show a) => Label a where label :: a -> String instance Label [Char] where label str = str instance Label Char where label c = [c] -- Default case instance Show a => Label a where label x = show x However, because the default case's class overlaps instance Label [Char] and instance Label Char, those types don't work with the label function. Is there a library function that provides this functionality? If not, is there a workaround to get the above code to work?

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  • strange error in haskell about indentation of if-then-else

    - by Drakosha
    I have the following code: foo :: Int -> [String] -> [(FilePath, Integer)] -> IO Int foo _ [] _ = return 4 foo _ _ [] = return 5 foo n nameREs pretendentFilesWithSizes = do result <- (bar n (head nameREs) pretendentFilesWithSizes) if result == 0 then return 0 -- <========================================== here is the error else foo n (tail nameREs) pretendentFilesWithSizes I get an error on the line with the comment above, the error is: aaa.hs:56:2: parse error (possibly incorrect indentation) I'm working with emacs, there's no spaces, and i do not understand what did i do wrong.

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  • The reason for MonadState get and put?

    - by CiscoIPPhone
    I'm reading the Monads chapter in Real World Haskell (chapter 14). A function is defined as follows: type RandomState a = State StdGen a getRandom :: Random a => RandomState a getRandom = get >>= \gen -> let (val, gen')= random gen in put gen' >> return val I don't really understand the purpose of the get and put functions here. I rewrote the function as following which seems to do the same thing and is more concise: getRandom2 :: Random a => RandomState a getRandom2= State $ \ s -> random s So my question is: What is the purpose of get and put?

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