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I have simple question about Haskell
by Eunsu Kim 23 Apr '16

23 Apr '16

Hi THIS IS MY CODE: evalpoly = do putStr "What is the degree of polynomial: " degree <- getLine coeffs <- (funcOfCoeff ((read degree::Int)+1) [] ) putStr "What value do you want to evaluate at: " value <- getLine putStr "the value of " putStr (printChar coeffs) putStr (" evaluated at "++ value ++" is ") putStrLn (show (getResult (coeffs) (read value :: Float) )) printChar coeffs =parser([x | x <-reverse (zip coeffs (iterate (+1) 0)), fst x /= 0]) parser []="" parser (a:as)=show(fst a) ++ "x^" ++show(snd a) ++ " " ++(parser as) —PROBLEM IS HERE!!! ---function loop to get coefficient-- funcOfCoeff 0 coeffs = do --to check the degree of 0 return coeffs --return list of coefficient funcOfCoeff degree coeffs = do putStr ("What is the x^" ++ show(degree-1)) putStr " coefficient: " coeff <- getLine loop <- funcOfCoeff (degree-1) ((read coeff :: Float) : coeffs) return loop getResult (coeffs) x = sum(map(\(a,b) -> a*x^b).zip coeffs.iterate (+1)$0) --evaluate polynomial with value HERE IS MY OUTPUT: *Main> evalpoly What is the degree of polynomial: 2 What is the x^2 coefficient: 3 What is the x^1 coefficient: 2 What is the x^0 coefficient: 1 What value do you want to evaluate at: 7 the value of 3.0x^2 2.0x^1 1.0x^0 evaluated at 7 is 162.0 in output, on the very bottom part, there should be + or - sign there like this: 3.0x^2 + 2.0x^1 - 1.0x^0 what should I do? I have no idea now….. thanks

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Haskell triangular loop (correct use of (++))
by rohan sumant 23 Apr '16

23 Apr '16

Suppose I have a list of distinct integers and I wish to generate all possible unordered pairs (a,b) where a/=b. Ex: [1,2,3,0] --> [(1,2),(1,3),(1,0),(2,3),(2,0),(3,0)] The approach I am following is this :- mkpairs [] = [] mkpairs (x:xs) = (map (fn x) xs) ++ (mkpairs xs) fn x y = (x,y) It is generating the desired output but I am a bit unsure about the time complexity of the function mkpairs. In an imperative language a nested triangular for loop would do the trick in O(n^2) or more precisely (n*(n-1)/2) operations. Does my code follow the same strategy? I am particularly worried about the (++) operator. I think that (++) wouldn't add to the time complexity since the initial code fragment (map (fn x) xs) is to be computed anyway. Am I wrong here? Is this implementation running O(n^2)? If not, could you please show me how to write a nested triangular loop in Haskell? Rohan Sumant

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I have question about Haskell
by Eunsu Kim 22 Apr '16

22 Apr '16

Hi i have a problem in my code! here is my code: -- Baic I/O and Loop (50 Points) evalpoly = do putStr "What is the degree of polynomial: " degree <- getLine coeffs <- (funcOfCoeff ((read degree::Int)+1) []) putStr "What value do you want to evaluate at: " value <- getLine putStr "The value of the polynomial is: " putStr (show (polyEvaluate (coeffs) (read value :: Float) )) putStr "\n" --function loop to get coefficient-- funcOfCoeff 0 coeffs = do --to check the degree of 0 return coeffs --return list of coefficient funcOfCoeff degree coeffs = do putStr ("What is the x^" ++ show(degree-1)) putStr " coefficient: " coeff <- getLine loop <- funcOfCoeff (degree-1) ((read coeff :: Int) : coeffs) return loop polyEvaluate (coeffs) x = do powers <- zip coeffs (iterate (+1) 0) result <- map (\(a,b)-> a+b) powers —PROBLEM IS HERE!!!! return result here is error message: in very bottom function (polyEvaluate), why is not working “result <- map (\(a,b) -> a+b) powers” ??? in Prelude, it is working Thanks!

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ghci :load vs import
by Doug McIlroy 19 Apr '16

19 Apr '16

I have module Powser stored in haskell/Powser.hs. There is no file ./Powser*. This loads the module shellprompt> ghci -ihaskell Prelude> :load Powser But this can't find it shellprompt> ghci -ihaskell Prelude> import Powser What might cause the difference? (I am running ghci 7.8.4.) Doug

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ghci :load vs import
by Doug McIlroy 19 Apr '16

19 Apr '16

I have module Powser stored in haskell/Powser.hs. There is no file ./Powser*. This loads the module shellprompt> ghci -ihaskell Prelude> :load Powser But this can't find it shellprompt> ghci -ihaskell Prelude> import Powser What might cause the difference? (I am running ghci 7.8.4.) Doug

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Parsec and Parsing
by mike h 18 Apr '16

18 Apr '16

Hi, I have two types, Tag = String Val = String and I want to creat expressions based on equality, t = v (where t is Tag, v is Val) and containment ie t IN v1, v2, v3, v4 ie a tag equals a value or a tag is contained in a list of values. Furthyermore I would like to apply AND and OR, i.e. (t1 = v OR t1 IN v1, v2, v3) AND (t2 = v2) ect etc I want to use Parsec and I'm ok with basic combinators but I want to get as far as making this a simple DSL but I don't want to use buildExpressionParser - at least not initially. I want to do this from first principles, understand what I'm doing and then maybe use buildExpressionParser. Any help would be really appreciated. Thanks Mike

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Trying to install "Craft3e" from Hackage using stack
by Albin Otterhäll 16 Apr '16

16 Apr '16

I'm currently reading "Haskell: the Craft of Functional Programming", by Simon Thompson and I'm trying to download and install the package books code. But when I run "stack build Craft3e" I get the following error message: > Run from outside a project, using implicit global project config > Using resolver: lts-5.12 from implicit global project's config file: > /home/dumbl3d0re/.stack/global-project/stack.yaml > While constructing the BuildPlan the following exceptions were encountered: > > -- Failure when adding dependencies: > HUnit: needed (==1.2.*), 1.3.1.1 found (latest applicable is 1.2.5.2) > mtl: needed (>=1.1 && <2.2), 2.2.1 found (latest applicable is 2.1.3.1) > needed for package Craft3e-0.1.0.10 How should I go about to install "Craft3e"? Thanks in advance! Regards, Albin

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Unicode characters in function names: some don't work?
by Silent Leaf 14 Apr '16

14 Apr '16

Hi! say I wanna use "—" as new infix operator: it's the big dash used a bit like parenthesis, especially at the end of sentences —like this. in ghci directly: Prelude> let (—) a b = a + b No problem, is accepted and usable. Same in files. Now I try using (»), a French (amongst others) punctuation sign, typically replaces the quote-ends, « like this ». Doesn't work: <interactive>:2:6: lexical error at character '\187' I thought Haskell was Unicode-friendly? Why some symbols but not others? :'(

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function application
by mike h 12 Apr '16

12 Apr '16

Hi, I’m looking at mc :: (Integral a) => a -> a mc x | x > 100 = x - 10 | otherwise = mc ( mc ( x + 11 ) ) the mc ( mc ( x + 11 ) ) can also be written as mc . mc $ x + 11 and I expected it could also be written as mc . mc ( x + 11 ) but the compiler error starts off with Couldn't match expected type ‘a’ with actual type ‘a0 -> c0’ so that is telling me, isn’t it (?) , that using parens is making the argument to the second mc into a function ‘a0 -> c0’ So is mc . mc $ x + 11 the only correct way to write this particular function in ‘.’ style ? Many Thanks Mike

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Re: [Haskell-beginners] Haskell triangular loop (correct use of (++))
by Silent Leaf 12 Apr '16

12 Apr '16

I'm sorry to hear, no implicit memoization (but then is there an explicit approach?). in a pure language, this seems hardly logical, considering the functional "to one output always the same result" and the language's propensity to use recursion that uses then previous values already calculated. Really hope for an explicit memoization! and i don't mean a manual one ^^ if that is even possible? Anyway, i just don't get your function f. you did get that in mine, xs was the value of my comprehension list, aka [.s1..n] ++ [0] >From there, if I'm not wrong, your function creates a list of all truncated lists, I supposed to be used with a call for the nth element? True, it memorizes all needed things, and in theory only "drops" once per element of the list. As for incorporating it, i'm thinking, local variable? ^^ the function itself could be outside the comprehension list, in a let or where, or completely out of everything, I don't really see the problem. then it's just a matter of it being called onto xs inside the comprehension list, like that: result = [(x,y) | let xs = [1,2,3,0], let yss = f xs, (x,i) <- zip xs [1,2..], y <- yss !! i, x /= y] The remaining possible issue is the call to !!... dunno if that's costy or not. The best way to go through the list I suppose would be by recursion... oh wait, I'm writing as I'm thinking, and I'm thinking: result = [(x,y) | let xs = [1,2,3,0], let yss = f xs, x <- xs, ys <- yss, y <- ys, x /= y] after all, why not use the invisible internal recursion? What do you think? As for the number-instead-of-number-list, the crucial point i mentioned was to determine the maximum number of digit (biggest power of ten reached), and fill up the holes of the smaller numbers with zeroes, so your examples would be: [1,2,3] = 123 --maximum size of a number = 1, no need to fill up [12,3] = 1203 --maximum size of a number = 2 digits, thus the hole beside 3 gets filled with a zero, just like on good old digital watches ^^ Do you think extraction of clusters of digits from numbers would be advantageous, efficiently speaking?

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