- Beginners - Haskell.org

Getting a variables type as a string at runtime
by Gareth Morgan 13 Jan '14

13 Jan '14

Bit of a simple question but I'm struggling to find information on this via Google. I have a sum type, when the input is type A I want to process it, when it is type B I want to return MyError "Expected type A but actually got type B". However I also have C,D,E,F, etc which should also return similar errors so don't want to hardcode this string. Is there a function I can call on a variable to get the name of the data constructor that created it? I could create a typeName function but I want to avoid boiler plate if I can. For reference what I'm actually doing is parsing Java class files. The constant pool I've represented as a map of a sum ConstantPoolEntry type (the Java constant pool for some reason skips an index on some types so can't be a simple list). Certain Java class file entries have a reference to the constant pool that must be of a specific constant type so I want to report badly formed entries to the user. Thanks, Gareth

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combine expressions in do block
by Renah Scarowsky 09 Jan '14

09 Jan '14

Is it possible to bind two expressions together within a case inside a do block? Given the following code running in the writer monad: f a = do case g a of Nothing -> return () Just b -> h a (field1 b) (field2 b) case g a of Nothing -> return () Just b -> i a (j a) (field1 b) (field2 b) How can I combine these two cases together (or otherwise simplify the code)? Thanks, Renah Scarowsky Suite Solutions Create>Manage>Deploy <http://www.suite-sol.com/> http://www.suite-sol.com

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Picking apart getLine
by Angus Comber 08 Jan '14

08 Jan '14

Before looking at getLine, I can understand this: getnumber :: IO Int getnumber = do x <- getChar if isDigit x then return (ord x - ord '0') else return 0 OK, it is not a very useful function but at least I understand it. return is required so that the function returns an IO Int. But I don't understand this: getLine' :: IO String getLine' = do x <- getChar if x == '\n' then return [] else do xs <- getLine' return (x:xs) I can understand what will happen if a user enters a newline (only). return [] brings the empty list into the monadic world. But what is happening if x is not a newline? xs <- getLine' will recursively call getChar and retrieve another character from the input stream. But it will do this BEFORE the return (x:xs) - so what is happening to all the head elements in the list - the x element? It is difficult to picture in my mind how this is working. I understand recursion but this looks tricky. Can someone help me work this out?

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Why WinGHCi does not complain with no import?
by Angus Comber 07 Jan '14

07 Jan '14

I created my own version of getLine like this: getLine' :: IO String getLine' = do x <- getChar if x == '\n' then return [] else do xs <- getLine' return (x:xs) Basically I created a new file called test.hs and copied above code into it. I opened WinGHCi afresh and :load test.hs I can run getLine' in WinGHCi with no problem. But how does WinGHCi know about getChar and IO? I assumed I would have to import like this: import System.IO Is it because WinGHCi does this import for me? If I write this program and invoke using runghci: getLine' :: IO String getLine' = do x <- getChar if x == '\n' then return [] else do xs <- getLine' return (x:xs) main = do getLine' It also works without complaining. Is there a list of built-in libraries? Is that platform dependent?

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Re: [Haskell-beginners] How to run Parser in chapter 8 Graham Hutton Haskell book
by Angus Comber 07 Jan '14

07 Jan '14

For anyone else benefit, this is how I fixed issue. 1. download Parsing.lhs from author's website. 2. Also download parser.lhs and paste contents into my prog_haskell.hs file, removing text comments and > symbols. 3. At top of my prog_haskell.hs file you need import Parsing You can then call the functions in WinGHCi, eg eval "3+5*2" For some reason I was not able to work out how to import both Parsing.lhs and parser.lhs together. Will have to read up on modules later. On 7 January 2014 11:43, Christian Maeder <Christian.Maeder(a)dfki.de> wrote: > Hi Angus, > > did you get an answer for your question? (Just pressing "reply" only > answers to the list, as happened to my earlier answer to your initial > question.) > > I don't know if WinGHCi works like ghci under linux (and I don't know > which files Parsing.lhs and parser.lhs you tried to load.) > > At the ghci Prompt you can type ":browse" to see which names are in scope. > Maybe parser.lhs could not be loaded for some reason that I cannot > reproduce. (Maybe eval is explicitly not exported?) > > Cheers Christian > > > Am 03.01.2014 18:43, schrieb Angus Comber: > >> I found some notes on the authors website explaining that the book code >> would no longer work. So I downloaded two files: Parsing.lhs and >> parser.lhs. In WinGHCi I then issued :load on both files. >> >> parser.lhs contains a function, eval, as follows: >> >> <other stuff above> >> > eval :: String -> Int >> > eval xs = case (parse expr xs) of >> > [(n,[])] -> n >> > [(_,out)] -> error ("unused input >> " ++ out) >> > [] -> error "invalid input" >> >> But in WinGHCi if I run: eval "2*3+4" I get error := Not in scope: `eval' >> >> I assume this is some sort of namespaces feature of Haskell that I have >> not read about? Or I somehow have to state what library modules to >> use??? Can anyone help? >> >> >> >> >> >> >> >> On 3 January 2014 15:58, Angus Comber <anguscomber(a)gmail.com >> <mailto:anguscomber@gmail.com>> wrote: >> >> I am reading Chapter 8 of Programming Haskell by Graham Hutton and >> trying to run the code in the book. >> >> It seems that there are already defined library implementations of >> Parser so I used Parser' and generally ' on the end of each >> identifier to attempt to eliminate this problem. >> >> So the code I copied from the book is: >> >> type Parser' a = String -> [(a, String)] >> >> return' :: a -> Parser' a >> return' v = \x -> [(v, x)] >> >> failure' :: Parser' a >> failure' = \inp -> [] >> >> -- item doesn't seem to conflict with anything so no ' >> item :: Parser' Char >> item = \inp -> case inp of >> [] -> [] >> (x:xs) -> [(x,xs)] >> >> >> parse' :: Parser' a -> String -> [(a, String)] >> parse' p inp = p inp >> >> >> p :: Parser' (Char, Char) >> p = do x <- item >> item >> y <- item >> return' (x,y) >> >> When run from WinGHCi I get error: >> >> prog_haskell.hs:458:9: >> Couldn't match type `[(Char, String)]' with `Char' >> Expected type: String -> [((Char, Char), String)] >> Actual type: Parser' ([(Char, String)], [(Char, String)]) >> In the return type of a call of return' >> In a stmt of a 'do' block: return' (x, y) >> In the expression: >> do { x <- item; >> item; >> y <- item; >> return' (x, y) } >> >> 458.9 is line at end with return' (x,y) >> >> >> Also in the chapter was the definition of >>= sequencing. as in: >> >> (>>=) :: Parser a -> (a -> Parser b) -> Parser b >> p >>= f = \inp -> case parse p inp of >> [] -> [] >> [(v,out)] -> parse (f v) out >> >> But I assumed this would already be in standard Haskell library and >> so I didn't need to define. But even when I did add, still get same >> error. >> >> How do I fix this? >> >> >> >> >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://www.haskell.org/mailman/listinfo/beginners >> >> >

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Using IORef
by Courtney Robinson 07 Jan '14

07 Jan '14

I've found this example. counter :: IO Int counter = unsafePerformIO $ newIORef 0 getUnique :: IO Int getUnique = atomicModifyIORef counter $ \x -> let y = x + 1 in (y, y) Which I ran, and I know works in as much as calling getUnique multiple times results in an incremented value. I have two things I'm failing to see about it. 1) Why doesn't counter return a newIORef with value 0, every time getUnique is called there by producing a value of 1 no matter how often getUnique is called? Is this this because it returns an IORef so counter just becomes bound to the value it first returns and it's not evaluated again and the IORef is just returned, or something else? 2) Why does the lambda return a tuple and why does the example return the same value in both positions i.e. (y,y)? The only thing I could think of is because the name has "atomic" that the values in the tuple is being used for some sort of compare and swap,CAS. P.S I didn't find http://hackage.haskell.org/package/base-4.6.0.1/docs/Data-IORef.html enlightening on the matter and I know that example is not great because the atomicModifyIORef is lazy so could lead to stackoverflow as that doc. page says... Thanks

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"Segmentation fault/access violation" when working with FFI
by Leza Morais Lutonda 06 Jan '14

06 Jan '14

Hi, I has trying to use the portaudio [1] package to play some sound, and running the examples [2] results in "Segmentation fault/access violation" error. How can I solve this problem? Thanks. [1] http://hackage.haskell.org/package/portaudio [2] https://github.com/sw17ch/portaudio/blob/master/examples/Example1.h

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Constrained polymorphic functions as natural transformations
by Matt R 06 Jan '14

06 Jan '14

I read this excellent blog article on natural transformations[1], which helped me gain one intuition as to what they are: "functions that don't make too many assumptions about their arguments". In the article, natural transformations make "no assumptions at all" about their arguments. However, it's often the case that a Haskell function will require a type argument constrained by a type class, e.g. show :: Show a => a -> String. I wondered if it was possible to characterise these functions using category theory too. Below is my attempt, and I'm hoping people can tell me whether I'm in the right ballpark (or point me in the direction of a better ball-park...) For a particular type class T, we can form a subcategory Hask_T of Hask consisting of all the instances of the type class as objects, with the arrows just those functions that "commute" with the signature of the type class. For example, in Hask_Show, the arrows would be all the functions f: A -> B such that: show == show . f Or for Data.Monoid, all the monoid homomorphisms f: mappend (f x) (f y) == f (mappend x y) mempty == f mempty In general, for any type class, we can formulate two functors f and g and a function sig :: f a -> g a that captures the operations in the type class, and then the condition is that: fmap f . sig == sig . fmap f. Then we have that the polymorphic functions with a type class constraint of T are just the same thing as natural transformations in this subcategory Hask_T. Is the above correct / along the right lines? Thanks, -- Matt [1] "You Could Have Defined Natural Transformations", http://blog.sigfpe.com/2008/05/you-could-have-defined-natural.html

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Help with Data.Aeson json data parsing
by Miro Karpis 04 Jan '14

04 Jan '14

Hi, please can you help me with following? I'm trying to parse one json (simplified version below), but I keep getting: "when expecting a record (:*:), encountered Array instead" code: ------------ data SomeData = SomeData { v1 :: Int, v2 :: Int } deriving (Show, Generic) data SomeDataPack = SomeDataPack{ pack :: [SomeData] } deriving (Show, Generic) instance FromJSON SomeData instance FromJSON SomeDataPack parsedata :: IO (Maybe SomeDataPack) parsedata = do let x = "[{\"v1\":11,\"v2\":12},{\"v1\":13,\"v2\":14}]"; case (eitherDecode' x :: Either String SomeDataPack) of Right r -> do putStrLn $ show r return $ Just r Left e -> do putStrLn $ show e return Nothing thanks, m.

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How to run Parser in chapter 8 Graham Hutton Haskell book
by Angus Comber 04 Jan '14

04 Jan '14

I am reading Chapter 8 of Programming Haskell by Graham Hutton and trying to run the code in the book. It seems that there are already defined library implementations of Parser so I used Parser' and generally ' on the end of each identifier to attempt to eliminate this problem. So the code I copied from the book is: type Parser' a = String -> [(a, String)] return' :: a -> Parser' a return' v = \x -> [(v, x)] failure' :: Parser' a failure' = \inp -> [] -- item doesn't seem to conflict with anything so no ' item :: Parser' Char item = \inp -> case inp of [] -> [] (x:xs) -> [(x,xs)] parse' :: Parser' a -> String -> [(a, String)] parse' p inp = p inp p :: Parser' (Char, Char) p = do x <- item item y <- item return' (x,y) When run from WinGHCi I get error: prog_haskell.hs:458:9: Couldn't match type `[(Char, String)]' with `Char' Expected type: String -> [((Char, Char), String)] Actual type: Parser' ([(Char, String)], [(Char, String)]) In the return type of a call of return' In a stmt of a 'do' block: return' (x, y) In the expression: do { x <- item; item; y <- item; return' (x, y) } 458.9 is line at end with return' (x,y) Also in the chapter was the definition of >>= sequencing. as in: (>>=) :: Parser a -> (a -> Parser b) -> Parser b p >>= f = \inp -> case parse p inp of [] -> [] [(v,out)] -> parse (f v) out But I assumed this would already be in standard Haskell library and so I didn't need to define. But even when I did add, still get same error. How do I fix this?

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