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Understanding functions like f a b c = c $ b a
by Austin Zhu 10 Aug '20

10 Aug '20

Hello! I'm learning Haskell and I found an interesting implementation of init using foldr. However I have difficulty understand how it works. *init' xs = foldr f (const []) xs id* * where f x g h = h $ g (x:)* Consider I have a input of *[1,2,3]*, then is would become *f 1 (f 2 ( f 3 (const []))) id* I substitute those parameters into f and the innermost one becomes *h $ (const []) (1:)*, which is simply *h []*. However when I want to reduce the expression further, I found it's hard to grasp. The next one becomes *f 2 (h [])* , which is *h $ (h []) (2:)* if it works like that. This looks confusing to me. To match the type of *foldr*, h should be of type *[a] -> [a]* and *h []* would just be of type *[a]*, which isn't applicable to *(2:)*. I also thought it in another way that *f x g* returns a function of type *([a] -> [a]) -> [a],* this kinda makes sense considering applying *id* afterwards. But then I realized I still don't know what this *h* is doing here. It looks like *h* conveys *g (x:)* from last time into the next application. Did I miss something when I think about doing fold with function as accumulator? I'd really appreciate if anyone could help me with this.

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Need help setting up development environment via ghcup
by Sudaraka Wijesinghe 28 Jul '20

28 Jul '20

Hi, I have successfully installed `ghc` (8.8.4), `cabal-install` (3.2.0.0) via `ghcup` (0.1.8) and created/build and several example projects without any issues. In one of the projects I want to create a binary distribution tarball. For this I tried using `runhaskell Setup.hs copy` (configure before that) at one of the steps, and got the following error. ``` Configuring my-project-0.1.0.0... Setup.hs: Encountered missing or private dependencies: optparse-applicative ==0.15.* ``` `optparse-applicative` is a dependency of my project and it was installed by `cabal build`. I figured out this is because `cabal-install` and `runhaskell` (or `ghc` or `ghc-pkg`) used different package DBs. - `cabal-install` -> ~/.cabal/store/ghc-8.8.4/ - `runhaskell` -> ~/.ghcup/ghc/8.8.4/lib/ghc-8.8.4/ Is there a was I can make both `cabal-install` and `runhaskell` use the same package DB (or use both of them)? [X-POST: https://www.reddit.com/r/haskellquestions/comments/hydge5/need_help_setting…]

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How to structure an application?
by Tilmann 01 Jul '20

01 Jul '20

Hi, I hope to get some advice on how to structure an application. So I acquire a handle early on that I use all over the app, but I don't want to pass the handle itself around, but wrap the handle with "commands" that a) make a nicer api and/or b) only allow specific usecases of the handle. I tried and failed to use MonadReader in a straightforward way and now I'm wondering what options there are. Looking forward to your feedback, Best, Tilmann module Main where import Control.Monad import Control.Monad.Reader import Graphics.UI.WX import System.IO -- imagine many more commands like this one ping :: (MonadReader Handle m, MonadIO m) => m () ping = do h <- ask liftIO $ hPutStrLn h "ping" main :: IO () main = do let h = stdout -- in the real app, this handle isn't stdout of course but opened separately start $ runReaderT wxApp h wxApp :: (MonadReader Handle m, MonadIO m) => m () wxApp = do ping -- this works, but I don't need it here.. liftIO $ do f <- frame [ ] timer f [ interval := 1000 -- , on command := hputStrLn h "ping" -- this is what I try to avoid -- , on command := ping -- of course, this doesn't work, but it would be so nice.. , enabled := True] return () -- Alternatively main2 :: IO () main2 = do let h = stdout start $ runReaderT wxApp2 (mkCommands h) wxApp2 :: (MonadReader Commands m, MonadIO m) => m () wxApp2 = do commands <- ask liftIO $ do f <- frame [ ] timer f [ interval := 1000 , on command := ping2 commands , enabled := True] return () data Commands = Commands { ping2 :: IO () -- .. many more } mkCommands :: Handle -> Commands mkCommands h = Commands (hPutStrLn h "ping")

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CSES programming problems at https://cses.fi/problemset/
by Doug McIlroy 30 Jun '20

30 Jun '20

> I'm currently stuck on the Two Knights problem. Having placed one knight on the board, in how many places can you put the other? Doug McIlroy

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CSES Two Sets problem at https://cses.fi/problemset/task/1092/
by Julian Ong 29 Jun '20

29 Jun '20

After the Two Knights problem, I went on this next problem which requires that you separate 1..n into two sets with the same sum if possible. Again my algorithm in Haskell works but is apparently too slow. It fails for CSES test inputs >= 26560 where a solution exists. I'm starting to wonder if Haskell is fundamentally too slow compared to other languages. From what I've read that shouldn't be the case though. For this problem it looks like it's doable in Python (I haven't tried that). Most of the fastest solutions for these problems seem to be written in C++. If there's anyone who's trying to solve these problems in Haskell (they're really fun by the way if you've never checked them out) and has solved this one (or Two Knights) and passed all the tests, I'd love to hear how you did it. Thanks. --- -- CSES - Two Sets-- Given 1..n, separate into two sets of equal sums and if possible list the elements of each set of a possible solution or output NO if not possible main :: IO ()main = do line <- getLine let n = read line :: Integer putStrLn $ solveN n -- Observe that sum [1..n] = n*(n+1)/2 so each set sum must be n*(n+1)/4 and so the set sum must be divisible by 4 for the separation to be possible-- Then the algorithm starts adding numbers from n down to 1 until the next number would make the sum exceed the required set sum-- At this point you add one more number, which will be the lowest number, to fill in the gap to complete set1. Set2 is then the other numbers.solveN :: Integer -> StringsolveN n | (n * (n+1) `mod` 4 /= 0) = "NO" | otherwise = "YES\n" ++ show set1_count ++ "\n" ++ (unwords $ map show set1_list) ++ "\n" ++ show set2_count ++ "\n" ++ (unwords $ map show set2_list) where set_sum = (n * (n+1)) `div` 4 set1_part1 = takeWhile (\x -> x*(n+1-x) + sum [0..(n-x)] < (n * (n+1)) `div` 4) [n, n-1..1] set1_part2 = set_sum - sum set1_part1 set1_list = set1_part1 ++ [set1_part2] set1_count = (toInteger . length) set1_list set2_list = [x | x <- [1..n], not (x `elem` set1_list)] set2_count = (toInteger . length) set2_list---- Julian

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Using output of head in data constuctor
by Josh Friedlander 28 Jun '20

28 Jun '20

Given the following code: module Log where import Control.Applicative data MessageType = Info | Warning | Error Int type TimeStamp = Int data LogMessage = LogMessage MessageType TimeStamp String | Unknown String I want to create a log parser like this: module LogAnalysis where import Log parseMessage :: String -> LogMessage parseMessage xs | length(words(xs)) < 3 = Unknown xs | notElem(head(words(xs)) ["I", "E", "W"]) = Unknown xs | otherwise = LogMessage Info 3 head(words(xs)) But GHC gives me "• Couldn't match type ‘[a0] -> a0’ with ‘[Char]’ Expected type: String Actual type: [a0] -> a0" So it thinks I am giving it the *function* head, when I would like to give it the output. How do I fix this? Thanks in advance,

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CSES programming problems at https://cses.fi/problemset/
by Julian Ong 27 Jun '20

27 Jun '20

Hi - I'm working through these problems using Haskell and was curious if anyone else is doing that. I'm currently stuck on the Two Knights problem. I have implemented a solution using recursion but it's not fast enough to pass the tests. Has anyone been able to solve this problem using Haskell? Looking for some optimization tips. I'm not sure if there's a better way to implement the recursive algorithm - is it doing unnecessary work? The problem requires that you calculate solutions for 1..n so you need to keep track of all intermediate values. The program fails the speed test for n=10000 -- it needs to complete in less than a second. I hold out hope that it's doable in Haskell but I can't figure it out. The algorithm uses the solution for the (k-1)x(k-1) board and adds the additional possibilities when you add a new left column and bottom row to make a kxk board. My current attempt looks like this:---- import Control.Monad main :: IO ()main = do line <- getLine let n = read line :: Integer putStr $ unlines $ map show $ reverse $ solveK n solveK :: Integer -> [Integer]solveK k | k == 1 = [0] | otherwise = (solveFrameK k + head (solveK (k-1))) : solveK (k-1) -- Returns list of knight moves in the upper right (k-1) x (k-1) portion of the board excluding the first column and first rowmoveKnightUR :: Integer -> (Integer, Integer) -> [(Integer, Integer)]moveKnightUR k (c, r) = do (c', r') <- [(c-1, r+2), (c+1, r+2), (c+2, r+1), (c+2, r-1), (c+1, r-2), (c-1, r-2), (c-2, r-1), (c-2, r+1)] guard (c' `elem` [2..k] && r' `elem` [2..k]) return (c', r') -- Returns list of left and bottom border squares for k x k board in (col, row) format with (1, 1) being the lower left squaregenBorder :: Integer -> [(Integer, Integer)]genBorder k = [(1, a) | a <- [1..k]] ++ [(a, 1) | a <- [2..k]] -- Formula for combinations C(n, r)combinations :: Integer -> Integer -> Integercombinations n r = product [1..n] `div` (product [1..(n-r)] * product [1..r]) -- Calculates additional number of two knight placements along the left and bottom border and from that border into the upper right (k-1) x (k-1) regionsolveFrameK :: Integer -> IntegersolveFrameK k | k == 1 = 0 | k == 2 = 6 | otherwise = ((combinations (2*k-1) 2) - 2) + (k-1) * (k-1) * (2*k-1) - sum (map (toInteger . length) (map (moveKnightUR k) (genBorder k)))---- Julian

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Why do i need to take out the list for this to work
by Alexander Chen 17 Jun '20

17 Jun '20

hi, assigment: make your own element function with the any function. --elem with any myElemAny :: Eq a => a -> [a] -> Bool myElemAny a = any (== a) --elem with any myElemAny' :: Eq a => a -> [a] -> Bool myElemAny' a [x]= any (== a) [x] myElemAny' compiles but throws an error because it has a non-exhaustive pattern. Could somebody tell me why the list gives the function grieveness? thanks,

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Easy to use XML parser
by Alexander Chen 17 Jun '20

17 Jun '20

Hi, For eductional/testing purposes I am looking for an .xml parser which can output strictly .csv. The pipeline goes from .xml -> .csv -> SQL table. The fat ones are already written in another language. Can you recommend a parser that can achieve this, has understandable documentation and is easy to use? thanks in advance. p.s. It would be great if you have personal experience with the library. best,

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foldr point free notation
by Alexander Chen 10 Jun '20

10 Jun '20

hi, --any function foldr myAny'' :: (a-> Bool) -> [a] -> Bool myAny'' f = foldr (\a b -> f a || b) False this is the foldr notions. How would i make this point free? best,

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