August 2016 - Beginners

Multiline code in GHCi
by Shishir Srivastava 29 Aug '17

29 Aug '17

Hi, I am trying to run this basic multiline code in GHCi. I am not sure if this is even possible without writing it in a file and compiling it in a traditional way. ------- Prelude> :set +m Prelude> let main1 = do Prelude| a <- readLn Prelude| return a Prelude| Prelude> Prelude> main1 <interactive>:75:1: No instance for (Show (IO b0)) arising from a use of `print' In the first argument of `print', namely `it' In a stmt of an interactive GHCi command: print it ------- Could someone please confirm if the function can be defined and called like this in GHCi or point to where am going wrong. Thanks, Shishir Srivastava

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GADTs and method signature
by PICCA Frederic-Emmanuel 06 Sep '16

06 Sep '16

Hello I have these types data ImageXY data Image3d' data ScPipeParams a where ScPipeParamsImageXY :: ScBitSize -> Channel -> Modulo -> Sc3du -> Roi -> Accumulation -> ScPipeParams ImageXY ScPipeParamsImage3d' :: ScBitSize -> Channel -> Modulo -> Sc3du -> Roi -> Accumulation -> ScPipeParams Image3d' and now I write a method scPipeOpen2' (ScDevDesc dev) p = alloca $ \params -> do poke params p res <- c_sc_pipe_open2 dev (scPipeType p) params checkError res PipeId scPipeOpen2 :: ScDevDesc -> ScPipeParams a -> IO (PipeId) scPipeOpen2 d p@(ScPipeParamsImageXY _ _ _ _ _ _) = scPipeOpen2' d p scPipeOpen2 d p@(ScPipeParamsImage3d' _ _ _ _ _ _) = scPipeOpen2' d p it works fine In order to avoid these _ _ _ _ _ _, I tryed to removed the @(Constructor ____) since I just pass the p parameter to the ' method. But in that case , I get this error message C:\Users\TEMPO\Downloads\tdc\tdc\srcLib.hsc:247:19: No instance for (Storable (ScPipeParams a)) arising from a use of scPipeOpen2' In the expression: scPipeOpen2' d p In an equation for `scPipeOpen2': scPipeOpen2 d p = scPipeOpen2' d p Indeed I already defined two instance of Storable instance Storable (ScPipeParams ImageXY) where ... and instance Storable (ScPipeParams Image3d') where ... Should I use a special extension in order to be able to write only scPipeOpen2 d p = scPipeOpen2' d p thanks for your help Frederic

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regex and Unicode
by Brian Sammon 30 Aug '16

30 Aug '16

I tried to write a program using Text.Regex.PCRE to search through a UTF8-encoded document. It appears that the presence of non-breaking-space characters (Charpoint 160) triggers some weird behavior in my program. This is using the Debian stable(Jessie) packages of ghc 7.6.3 and libraries. Now I find myself at a fork in the road, not sure which direction to head in. Do I: 1) Continue looking (or get help with looking) for bugs in my code? (I have this reduced to a pretty small test case) 2) Assemble a bug-report against debian? 3) Assemble a bug-report against Text.Regex.PCRE (or Text.Regex.Base) for "upstream" 4) Uninstall Text.Regex.PCRE (and/or some other packages) and switch to something that works with Unicode/UTF8? Any ideas?

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Learning Monads with 'setjmp' and 'longjmp' like actions
by Michael Roth 30 Aug '16

30 Aug '16

Hi, I'm really trying hard to understand monads but I guess my brain is trapped in the imperative world. Below what I have done so far. But now I would like to implement something like this: do foo (label, jumped) <- setjmp bar longjmp label But I'm stuck. How can I implement 'setjmp' and 'longjmp' with my monad? I'm not able to grasp how can I design 'setjmp' and somehow create a label to use in 'longjmp'. Maybe my monad structure is not appropriate? Code follows: --------------------------------------------------------------- data World = SomeWorld deriving (Eq, Show) data ProgramResult a = Result { world :: World, result :: a } | Continue { world :: World, continue :: Program a } | Stopped { world :: World } newtype Program a = Program { runProgram :: World -> ProgramResult a } runComplete :: Program a -> World -> (World, a) runComplete prg world0 = case runProgram prg world0 of (Result world1 x1) -> (world1, x1) (Continue world1 prg1) -> runComplete prg1 world1 nop :: Program () nop = return () yield :: Program () yield = Program $ \world -> Continue world nop stop :: Program () stop = Program $ \world -> Stopped world instance Functor Program where fmap = liftM instance Applicative Program where pure = return (<*>) = ap instance Monad Program where return x = Program $ \world0 -> Result world0 x ma >>= fb = Program $ \world0 -> case runProgram ma world0 of (Result world1 x1) -> runProgram (fb x1) world1 (Continue world1 prg1) -> Continue world1 (prg1 >>= fb) (Stopped world1) -> Stopped world1

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Could anyone explain what this means? (Haskell wiki entry on monads)
by umptious 30 Aug '16

30 Aug '16

>>Each monad, or computation type<< That seems to be saying that monad and "computation type" are synonyms. But surely this is just bad writing? A monad is a CT, but a CT doesn't have to be a monad? And what is the point of adding the verbiage of "computation type" in what will be a massive sentence anyway? >>provides means, subject to *Monad Laws*,<< I think it's expected that monads will follow monad laws and provide means to do something.. >>> to *(a)* *create* a description of computation action that will produce (a.k.a. "return") a given Haskell value<<< This is just bizarre. The monad can create a description of a computation action? What does that mean - a comment is a description - does it mean that? And does it mean the description will do the returning or that the computation action will? And why say "means" instead of something more specific and meaningful (for example, I'm guessing the means might be a function..)

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Why is the type constraint different from signature?
by Lai Boon Hui 29 Aug '16

29 Aug '16

Hi All, can some one explain to me why ghci> let f:: (Ord a, Num b) => a -> b -> a ; f=undefined ghci> :t f 1 2 ghci> f 1 2 :: (Num a, Ord a) => a The initial type signature just required *a* to be a type that is an instance of Ord but after it had the additional constraint of Ord as well??? -- Best Regards, Boon Hui

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How do you show a Data.ByteArray?
by haskell＠verge.info.tm 28 Aug '16

28 Aug '16

Prelude Data.ByteArray> c = Data.ByteArray.zero 20 Prelude Data.ByteArray> :type c c :: ByteArray ba => ba Prelude Data.ByteArray> c <interactive>:38:1: error: • Ambiguous type variable ‘a0’ arising from a use of ‘print’ Uhm... Prelude Data.ByteArray> index c 3 <interactive>:39:1: error: • Ambiguous type variable ‘a0’ arising from a use of ‘index’ I can't print a byte?

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Basic Trie Implementation: Request for Feedback and QuickCheck Question
by Dominik Bollmann 28 Aug '16

28 Aug '16

Hello Haskellers, I just finished implementing a very simple Trie module, consisting of a Trie data type as well as operations `insert` and `lookup`. Function `insert` inserts a new element into the Trie; function lookup searches for all words in the Trie that match a specific pattern. Besides regular words, a pattern may consist of a single dot '.', indicating that at this position of the search term any letter may occur. For example a pattern "hello" would just search for the word "hello" in the Trie, while ".ello" would search words starting in *any* letter and followed by the sequence "ello" (i.e., it searches for "hello", "Hello", "Aello", etc.) The code for the Trie module can be found here: http://lpaste.net/180768. Regarding the code I have two questions: (1) Since I'm new to Haskell, I'd very much welcome feedback on the general implementation of the Trie module and its two methods what I could improve. (2) While testing the Trie module with QuickCheck, I wrote the property `prop_insert_then_lookup_succeeds` stating that after inserting a word into a Trie it can always successfully be looked up again. While quickchecking this property seems to succeed, I cannot verbose check it. In particular, when I run `verboseCheck prop_insert_then_lookup_succeeds` this property consumes all my main memory and results in the program hanging. Does anyone know what I messed up in devising the quickcheck property? I guess I left open some memory leak... Maybe it has to to with the Arbitrary instance which I might have defined incorrectly? Responses to any of the two questions above are very welcome :-) Thanks! Dominik.

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cabal install hakyll: global constraint requires installed instance + other conflicts
by Stanislaw Findeisen 25 Aug '16

25 Aug '16

Hi I am trying to install hakyll in a cabal sandbox, but somehow it doesn't work. Here's what I am doing: > $ cabal sandbox init > $ cabal install hakyll > Resolving dependencies... > cabal: Could not resolve dependencies: > trying: hakyll-4.8.3.2 (user goal) > next goal: base (dependency of hakyll-4.8.3.2) > rejecting: base-4.6.0.1/installed-8aa... (conflict: hakyll => base>=4.8 && <5) > rejecting: base-4.9.0.0, 4.8.2.0, 4.8.1.0, 4.8.0.0, 4.7.0.2, 4.7.0.1, 4.7.0.0, > 4.6.0.1, 4.6.0.0, 4.5.1.0, 4.5.0.0, 4.4.1.0, 4.4.0.0, 4.3.1.0, 4.3.0.0, > 4.2.0.2, 4.2.0.1, 4.2.0.0, 4.1.0.0, 4.0.0.0, 3.0.3.2, 3.0.3.1 (global > constraint requires installed instance) > Backjump limit reached (change with --max-backjumps). > > Note: when using a sandbox, all packages are required to have consistent > dependencies. Try reinstalling/unregistering the offending packages or > recreating the sandbox. Then I tried (inside the sandbox): > $ cabal install cabal-install which installed cabal 1.24.0.0 into the sandbox. Now, when I am using it, I am getting different errors: > $ .cabal-sandbox/bin/cabal install hakyll > Warning: cannot determine version of /usr/lib/ghc/bin/haddock-ghc-7.6.3 : > "" > Resolving dependencies... > cabal: Could not resolve dependencies: > trying: parsec-3.1.11/installed-4f0... (user goal) > trying: pandoc-1.17.2 (dependency of hakyll-4.7.5.2) > next goal: cmark (dependency of pandoc-1.17.2) > rejecting: cmark-0.5.3.1, cmark-0.5.2.1, cmark-0.5.2, cmark-0.5.1, cmark-0.5.0 > (conflict: parsec => text==0.11.3.1/installed-e38..., cmark => text>=1.0 && > <1.3) > rejecting: cmark-0.4.1, cmark-0.4.0.1, cmark-0.3.4, cmark-0.3.3.1, > cmark-0.3.3, cmark-0.3.2, cmark-0.3.1, cmark-0.3.0.1, cmark-0.3, > cmark-0.2.0.2, cmark-0.2.0.1, cmark-0.2, cmark-0.1.0.1, cmark-0.1.0.0 > (conflict: pandoc => cmark>=0.5 && <0.6) > rejecting: cmark-0.5.3 (conflict: parsec => text==0.11.3.1/installed-e38..., > cmark => text>=1.0 && <1.3) > rejecting: cmark-0.3.5 (conflict: pandoc => cmark>=0.5 && <0.6) > Backjump limit reached (currently 2000, change with --max-backjumps or try to > run with --reorder-goals). > > Note: when using a sandbox, all packages are required to have consistent > dependencies. Try reinstalling/unregistering the offending packages or > recreating the sandbox. What is going on here? Why am I getting different errors with the new cabal version? What is the relationship between system-wide packages and those in the sandbox? I thought sandbox is meant to be an isolated environment?!? How do I get the list of all the packages currently installed in the sandbox? Is this: $ ls ./lib/x86_64-linux-ghc-7.6.3/ async-2.1.0 ed25519-0.0.5.0 parsec-3.1.11 base16-bytestring-0.1.1.6 hackage-security-0.5.2.1 random-1.1 base64-bytestring-1.0.0.1 HTTP-4000.3.3 stm-2.4.4.1 bytestring-builder-0.10.8.1.0 mtl-2.2.1 tar-0.5.0.3 Cabal-1.24.0.0 network-2.6.3.1 transformers-0.5.2.0 cryptohash-sha256-0.11.100.1 network-uri-2.6.1.0 zlib-0.6.1.1 the way? Why is 'base' not on the list, although the new cabal is not complaining about it anymore? This is the latest Debian stable. Thanks! -- http://people.eisenbits.com/~stf/ http://www.eisenbits.com/ OpenPGP: 9EC2 5620 2355 B1DC 4A8F 8C79 0EC7 C214 E5AE 3B4E

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Functions as Applicatives
by Olumide 23 Aug '16

23 Aug '16

Hi List, I'm struggling to relate the definition of a function as a function instance Applicative ((->) r) where pure x = (\_ -> x) f <*> g = \x -> f x (g x) with the following expression ghci> :t (+) <$> (+3) <*> (*100) (+) <$> (+3) <*> (*100) :: (Num a) => a -> a ghci> (+) <$> (+3) <*> (*100) $ 5 508 From chapter 11 of LYH http://goo.gl/7kl2TM . I understand the explanation in the book: "we're making a function that will use + on the results of (+3) and (*100) and return that. To demonstrate on a real example, when we did (+) <$> (+3) <*> (*100) $ 5, the 5 first got applied to (+3) and (*100), resulting in 8 and 500. Then, + gets called with 8 and 500, resulting in 508." The problem is that I can't relate that explanation with the definition of a function as an applicative; especially f <*> g = \x -> f x (g x) . Is (g x) the second argument to f? Regards, - Olumide

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