December 2015 - Beginners

basic State Monad
by Imants Cekusins 15 Dec '15

15 Dec '15

This snippet increments integer n times using state monad. How to call: main 10 5 module BasicState where import Control.Monad.State.Strict import Debug.Trace type St a = State a a -- caller is not aware that main uses state -- mai is a pure function main :: Int -> Int -> Int main start0 repeat0 = evalState (repeatN repeat0) start0 -- state-passing computation repeatN :: Int -> St Int repeatN n0 -- repeat n times | n0 < 1 = get -- current state | otherwise = do withState pureStateModifier get -- update state repeatN $ n0 - 1 -- recurse -- state unaware modifier function pureStateModifier :: Int -> Int pureStateModifier = (+ 1)

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Understanding Haskell Map.lookup Example in LYH
by Olumide 14 Dec '15

14 Dec '15

Hello Haskellers! I'd appreciate help understanding the origin of the extra 'map' in 'Map.lookup lockerNumber map', from the following example taken from chapter 8 of "Learn You a Haskell for Great Good" http://learnyouahaskell.com/making-our-own-types-and-typeclasses#type-synon… lockerLookup :: Int -> LockerMap -> Either String Code lockerLookup lockerNumber map = case Map.lookup lockerNumber map of Nothing -> Left $ "Locker number " ++ show lockerNumber ++ " doesn't exist!" Just (state, code) -> if state /= Taken then Right code else Left $ "Locker " ++ show lockerNumber ++ " is already taken!" Regards, - Olumide

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Haskell code optimisation
by Abhishek Kumar 13 Dec '15

13 Dec '15

I was trying to write below program for ackerman function but it fails (waits too long) for ack(4,1) whereas a recursive C program gives result in 37secs.Can someone pls explain this behaviour and recomend some optimisation. ------haskell code f m n | m==0 =n+1 | n==0 = f (m-1) 1 | otherwise = f (m-1) (f m (n-1)) Thanks Abhishek Kumar

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Trouble understanding the type of sequence [Just, Just]
by Lim H. 12 Dec '15

12 Dec '15

Hi everyone, Sorry if this email disturbs you. I haven't used a developer's mailing list before so I'm not sure if I'm violating any etiquette. If I do, please excuse me. I'm trying to understand the type of sequence [Just, Just]. I can understand sequence [Just 1, Just 2] :: Num a => Maybe [a] because when looking at the type of sequence sequence :: (Monad m, Traversable t) => t (m a) -> m (t a) it is clear that this function takes a collection of monadic values and return a single monadic value of the collection. Thus, when we call sequence [Just 1, Just 2] we should get back a Just of [1,2]. Following that train of thoughts, shouldn't sequence [Just, Just] return a single Just? Here is the corresponding SO question http://stackoverflow.com/questions/34244574/trouble-understanding-the-type-… Lim

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foldr, Foldable and right-side folding
by Raja 11 Dec '15

11 Dec '15

foldr is supposed to start folding from the right side (as the name suggests). and this is why it is synonymous to "list construction" as I'm told for e.g: > foldr (:) [ ] [1,2,3,4,5] [1,2,3,4,5] In the same spirit I'm trying to construct a Foldable instance of my own type: data Li a = Nil | Cons a (Li a) deriving (Show) instance Foldable Li where foldr f b Nil = b foldr f b (Cons a y) = foldr f (f a b) y So I'm trying out foldr for my type: > foldr Cons Nil (Cons 1 (Cons 2 Nil)) Cons 2 (Cons 1 Nil) This shows my foldr implementation i'm not folding from right side, but how can I possibly do that - the data could have been an infinite stream. It feels like I will never be able to truly write a foldr implementation with "right" folding mechanism. Any thoughts?

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Help with expressing polymorphic return type
by Marco Yuen 11 Dec '15

11 Dec '15

Hi all, {-# LANGUAGE TypeFamilies #-} data CB = CB deriving (Show) data CO = CO deriving (Show) data CBParseResult = CBParseResult deriving (Show) data COParseResult = COParseResult deriving (Show) class ParseResult a where type EntityType a :: * toEntity :: a -> EntityType a instance ParseResult CBParseResult where type EntityType CBParseResult = CB toEntity result = CB instance ParseResult COParseResult where type EntityType COParseResult = CO toEntity result = CO getParseResult :: (ParseResult r) => String -> IO r getParseResult i = do if someCond then return CBParseResult else return COParseResult where someCond = null i Couldn't match expected type ‘r’ with actual type ‘COParseResult’ ‘r’ is a rigid type variable bound by the type signature for getParseResult :: ParseResult r => String -> IO r at Parse.hs:19:19 Relevant bindings include getParseResult :: String -> IO r (bound at Parse.hs:20:1) In the first argument of ‘return’, namely ‘COParseResult’ In the expression: return COParseResult What I want to express is getParseResult to be able to return any instances of ParseResult type class. But what I understand from the error is that r has been specialized to CBParseResult, but in the next expression I'm returning COParseResult. Is my understanding correct? And, How would I express what I described? Thanks, Marco

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Re: [Haskell-beginners] How can I lift my function taking a function to a Monad?
by Simon Peter Nicholls 11 Dec '15

11 Dec '15

Hiya, That would certainly explain why I found it difficult. Here's the original definition: liftMLabel :: Monad m => a :-> b -> m a :-> m b liftMLabel l = label (liftM $ get l) (liftM2 $ set l) Are here's a sample iteration of me hacking around: liftMLens :: Monad m => a :-> b -> m a :-> m b liftMLens l = lens getter modifier where getter = liftM $ get l modifier mf mm = mf >>= \ f -> modify l f `liftM` mm I only ever seem to replicate liftM2 characteristics when trying to brute force an implementation. And the implementations get more and more brutish as time goes on! It makes me wonder if the implications of fclabels `m a :-> m b` are radically different now, bearing in mind the original code worked. The fclabels :-> type operator has convinced the compiler that `m b -> mb` is required for the actual value modifying function. Does `m a :-> m b` make sense for the latest releases of fclabels? Is it just the lifting that's a problem? Si On Thu, 10 Dec, 2015 at 3:02 PM, Daniel Bergey <bergey(a)teallabs.org> wrote: > On 2015-12-10 at 07:43, Simon Peter Nicholls <simon(a)mintsource.org> > wrote: >> So I figure I need: >> >> neededLift :: (Monad m) => ((b -> b) -> a -> a) -> (m b -> m b) -> >> m a >> -> m a > > I don't believe `neededLift` is possible. The type says that given > any > function `m b -> m b`, you can turn that function into one of type `b > -> > b` (as input to the lens).

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problem with SDL-gfx
by Martín Villagra 10 Dec '15

10 Dec '15

Hello! I hope you can help me. I'm trying to use SDL-gfx but it seems it isn't linking it correctly. When I try to run this simple file with runhaskell: import Graphics.UI.SDL.Primitives main = putStrLn "Hello" It gives: sdl.test.hs: <command line>: can't load .so/.DLL for: /home/username/.cabal/lib/i386-linux-ghc-7.10.2/SDL-gfx-0.6.0.1-6rhSP5mw8M00K9S8yASTVF/ libHSSDL-gfx-0.6.0.1-6rhSP5mw8M00K9S8yASTVF-ghc7.10.2.so (/home/username/.cabal/lib/i386-linux-ghc-7.10.2/SDL-gfx-0.6.0.1-6rhSP5mw8M00K9S8yASTVF/ libHSSDL-gfx-0.6.0.1-6rhSP5mw8M00K9S8yASTVF-ghc7.10.2.so: undefined symbol: SDL_initFramerate) Notes: * SDL works fine in all cases. * My system is i386, using Arch Linux. I tried with another computer with x86_64 with Arch Linux and it worked. * It works if I compile it with ghc and run the executable. * I installed the package using cabal install SDL-gfx with no errors. * I installed the library in the OS just in case, and double checked that it exist in /usr/lib/libSDL_gfx.so and that it contains the symbol SDL_initFramerate. * The long *.so library reported by runhaskell also constains SDL_initFramerate. * GHCi gives the same error (when executing main). * I'm trying to make a cabal package that uses this library, and cabal gave the same error within the sandbox.

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How can I lift my function taking a function to a Monad?
by Simon Peter Nicholls 10 Dec '15

10 Dec '15

Hi All, Can anyone help me lift a function that takes a function, so it can be used in a Monad? The function I am given looks like "((b -> b) -> a -> a)", and viewing the definition of liftM2, it's almost what I need: liftM2 :: (Monad m) => (a1 -> a2 -> r) -> m a1 -> m a2 -> m r liftM2 f m1 m2 = do x1 <- m1 x2 <- m2 return (f x1 x2) but I get the error: Expected type: (m b -> m b) -> m a -> m a Actual type: m (b -> b) -> m a -> m a which makes sense, as liftM2 just wraps my function arg, whereas the function itself needs to operate on monadic values. So I figure I need: neededLift :: (Monad m) => ((b -> b) -> a -> a) -> (m b -> m b) -> m a -> m a Despite a rough appreciation of how liftM2 works ("unpack" desired monadic versions of the args, apply the function, then return to put result back in the Monad), my mind keeps hazing over regarding the changes I need to make in this case. I've tried implementing neededLift by composing other functions, and also by manual Monad wrangling, but the intuitive leap remains elusive. I posted the specific problem over at SO (http://stackoverflow.com/questions/34152747/how-can-i-lift-an-fclabels-lens…) but I think the specifics may be clouding what is probably a straightforward problem. Cheers, Si

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More type errors I'm having trouble with
by Dan Stromberg 09 Dec '15

09 Dec '15

I'm continuing my now-and-then exploration of Haskell. I'm getting a new crop of type errors that I'm pulling my hair out over. The errors I'm getting are: $ make below cmd output started 2015 Tue Dec 01 04:05:17 PM PST # --make will go out and find what to build ghc -Wall --make -o dph dph.hs Split0.hs [1 of 3] Compiling Split0 ( Split0.hs, Split0.o ) [2 of 3] Compiling Md5s ( Md5s.hs, Md5s.o ) [3 of 3] Compiling Main ( dph.hs, dph.o ) dph.hs:13:13: Couldn't match type `IO' with `(,) (IO String)' Expected type: (IO String, String) Actual type: IO String In a stmt of a 'do' block: hash <- prefix_md5 filename :: IO String In the expression: do { hash <- prefix_md5 filename :: IO String; (hash, filename) } In an equation for `do_prefix_hash': do_prefix_hash filename = do { hash <- prefix_md5 filename :: IO String; (hash, filename) } dph.hs:14:6: Couldn't match type `[Char]' with `IO String' Expected type: IO String Actual type: String In the expression: hash In a stmt of a 'do' block: (hash, filename) In the expression: do { hash <- prefix_md5 filename :: IO String; (hash, filename) } dph.hs:24:23: Couldn't match type `[]' with `IO' Expected type: IO (IO String, String) Actual type: [(IO String, String)] In a stmt of a 'do' block: io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)] In the expression: do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; hash_tuples <- sequence io_hash_tuples :: [(String, String)]; .... } In an equation for `main': main = do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; .... } dph.hs:25:20: Couldn't match type `[a0]' with `(String, String)' Expected type: [(String, String)] Actual type: [[a0]] In the return type of a call of `sequence' In a stmt of a 'do' block: hash_tuples <- sequence io_hash_tuples :: [(String, String)] In the expression: do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; hash_tuples <- sequence io_hash_tuples :: [(String, String)]; .... } dph.hs:25:20: Couldn't match type `[]' with `IO' Expected type: IO (String, String) Actual type: [(String, String)] In a stmt of a 'do' block: hash_tuples <- sequence io_hash_tuples :: [(String, String)] In the expression: do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; hash_tuples <- sequence io_hash_tuples :: [(String, String)]; .... } In an equation for `main': main = do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; .... } dph.hs:25:29: Couldn't match expected type `[[a0]]' with actual type `(IO String, String)' In the first argument of `sequence', namely `io_hash_tuples' In a stmt of a 'do' block: hash_tuples <- sequence io_hash_tuples :: [(String, String)] In the expression: do { buffer <- (hGetContents stdin) :: IO String; let filenames = ...; io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)]; hash_tuples <- sequence io_hash_tuples :: [(String, String)]; .... } dph.hs:26:39: Couldn't match expected type `[(String, String)]' with actual type `(String, String)' In the second argument of `map', namely `hash_tuples' In the expression: map tuple_to_string hash_tuples :: [String] In an equation for `strings': strings = map tuple_to_string hash_tuples :: [String] make: *** [dph] Error 1 above cmd output done 2015 Tue Dec 01 04:05:18 PM PST dph.hs looks like: import Md5s import Split0 import System.IO get_filenames :: String -> [String] get_filenames buffer = do -- Let's hope this doesn't give locale-related roundtrip problems. Split0.split0 '\0' buffer :: [String] do_prefix_hash :: String -> (IO String, String) do_prefix_hash filename = do hash <- Md5s.prefix_md5 filename :: (IO String) (hash, filename) tuple_to_string :: (String, String) -> String tuple_to_string (first, second) = do (show first) ++ " " ++ (show second) main :: IO () main = do buffer <- (System.IO.hGetContents System.IO.stdin) :: IO String let filenames = (get_filenames buffer) :: [String] io_hash_tuples <- map do_prefix_hash filenames :: [(IO String, String)] hash_tuples <- sequence io_hash_tuples :: [(String, String)] let strings = map tuple_to_string hash_tuples :: [String] mapM_ putStrLn strings And Md5s.hs looks like: module Md5s where import qualified System.IO import qualified Text.Printf -- cabal install cryptohash import qualified Crypto.Hash.MD5 import qualified Data.ByteString import qualified Data.ByteString.Lazy -- http://stackoverflow.com/questions/10099921/efficiently-turn-a-bytestring-i… byte_string_to_hex :: Data.ByteString.ByteString -> String byte_string_to_hex = concatMap (Text.Printf.printf "%02x") . Data.ByteString.unpack prefix_md5 :: String -> IO String prefix_md5 filename = do let prefix_length = 1024 file <- System.IO.openBinaryFile filename System.IO.ReadMode :: IO System.IO.Handle data_read <- Data.ByteString.hGet file prefix_length :: IO Data.ByteString.ByteString _ <- System.IO.hClose file let hasher = Crypto.Hash.MD5.init :: Crypto.Hash.MD5.Ctx let hasher2 = Crypto.Hash.MD5.update hasher data_read :: Crypto.Hash.MD5.Ctx let binary_digest = Crypto.Hash.MD5.finalize hasher2 :: Data.ByteString.ByteString let hex_digest = byte_string_to_hex binary_digest :: String return hex_digest :: IO String full_md5 :: String -> IO String full_md5 filename = do file <- System.IO.openBinaryFile filename System.IO.ReadMode :: IO System.IO.Handle data_read <- Data.ByteString.Lazy.hGetContents file :: IO Data.ByteString.Lazy.ByteString let binary_digest = Crypto.Hash.MD5.hashlazy data_read :: Data.ByteString.ByteString let hex_digest = byte_string_to_hex binary_digest :: String -- Does this get closed for us later? -- strace shows the file getting closed without our explicit close. -- _ <- System.IO.hClose file return hex_digest :: IO String It might be easier to view these at http://stromberg.dnsalias.org/svn/equivalence-classes/trunk/equivs3-haskell/ , so the line numbers are precise. What is the deal? Can anyone tell me what should be running through my head to fix this kind of problem on my own in the future? Thanks! -- Dan Stromberg

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