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Named fields in data types
by legajid 18 Nov '09

18 Nov '09

Hello, i've some trouble with named fields in data types. data Carnet = Adresse { nom :: String, cp :: Integer, ville :: String } | Telephone {nom::String, telnum::String} deriving Show contact01 = Adresse "Didier" 51100 "Reims" contact02 = Adresse {nom="Laure", cp=0} -- line 1 --contact02 Adresse { ville="Nogent" } -- line 2 --contact02 { ville="Nogent" } -- line 3 contact03=Telephone "Didier" "0326..." -- line 4 When loading this source : line 1 says "warning : Fields not initialized :ville line 2 and 3 when uncommented give parse error (possibly indentation) I'm ok with line 1. Is it possible to write such things as line2 or 3 ? Which syntax ? Thanks for help. Didier.

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Question on data/type
by Phillip Pirrip 18 Nov '09

18 Nov '09

Hi, I have the following data defined. data TypeCon a = ValConA a | ValConB [a] | ValConC [[a]] So I can use functions with type like (a->a->a) -> TypeCon a -> TypeCon a -> TypeCon a for all 3 value types, and I think is easier to define one single typeclass for (+), (*) etc. If I want to express the following idea (the following won't compiler): data TypeCon a = ValConA a | ValConB [ValConA a] | ValConC [ValConB a] Is this even a good idea? If so how could I proceed? The closest thing I can get to compiler is like this: data TypeCon a = ValConA a | ValConB [TypeCon a] Which is a nightmare when I try to manipulate anything in this structure. The alternative I guess is to use 3 different type constructors, data TypeConA a = ValConA a data TypeConB a = ValConB [ValConA a] data TypeConC a = ValConC [ValConB a] but then I can't use one signal typeclass for (+) etc. Am I correct? thx, //pip

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Questions on data/type
by Matthew Wong 17 Nov '09

17 Nov '09

Hi, I have the following data defined. data TypeCon a = ValConA a | ValConB [a] | ValConC [[a]] So I can use functions with type like (a->a->a) -> TypeCon a -> TypeCon a -> TypeCon a for all 3 value types, and I think is easier to define one single typeclass for (+), (*) etc. If I want to express the following idea (the following won't compiler): data TypeCon a = ValConA a | ValConB [ValConA a] | ValConC [ValConB a] Is this even a good idea? If so how could I proceed? The closest thing I can get to compiler is like this: data TypeCon a = ValConA a | ValConB [TypeCon a] Which is a nightmare when I try to manipulate anything in this structure. The alternative is use 3 different type constructors, data TypeConA a = ValConA a data TypeConB a = ValConB a data TypeConC a = ValConC a but then I can't use one signal typeclass for (+) etc. thx, //pip

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Functor question
by Matthew Wong 17 Nov '09

17 Nov '09

Hi, When I define my data as fellow, data Moo a = Moo a deriving (Show) instance Functor Moo where fmap f (Moo a) = Moo (f a) GHC gives me no problem. Then I add something, data (Num a) => Moo a = Moo a deriving (Show) instance Functor Moo where fmap f (Moo a) = Moo (f a) Now GHC gives me the follow error - see bellow. What is the reason behind this? What should I do to correct this? thx, // matFun_v2.hs:16:12: Could not deduce (Num a) from the context () arising from a use of `Moo' at matFun_v2.hs:16:12-14 Possible fix: add (Num a) to the context of the type signature for `fmap' In the pattern: Moo a In the definition of `fmap': fmap f (Moo a) = Moo (f a) In the instance declaration for `Functor Moo' matFun_v2.hs:16:21: Could not deduce (Num b) from the context () arising from a use of `Moo' at matFun_v2.hs:16:21-29 Possible fix: add (Num b) to the context of the type signature for `fmap' In the expression: Moo (f a) In the definition of `fmap': fmap f (Moo a) = Moo (f a) In the instance declaration for `Functor Moo'

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Functor question.
by Phillip Pirrip 16 Nov '09

16 Nov '09

Hi, When I define my data as fellow, data Moo a = Moo a deriving (Show) instance Functor Moo where fmap f (Moo a) = Moo (f a) GHC gives me no problem. Then I add something, data (Num a) => Moo a = Moo a deriving (Show) instance Functor Moo where fmap f (Moo a) = Moo (f a) Now GHC gives me the follow error - see bellow. What is the reason behind this? What should I do to correct this? thx, // matFun_v2.hs:16:12: Could not deduce (Num a) from the context () arising from a use of `Moo' at matFun_v2.hs:16:12-14 Possible fix: add (Num a) to the context of the type signature for `fmap' In the pattern: Moo a In the definition of `fmap': fmap f (Moo a) = Moo (f a) In the instance declaration for `Functor Moo' matFun_v2.hs:16:21: Could not deduce (Num b) from the context () arising from a use of `Moo' at matFun_v2.hs:16:21-29 Possible fix: add (Num b) to the context of the type signature for `fmap' In the expression: Moo (f a) In the definition of `fmap': fmap f (Moo a) = Moo (f a) In the instance declaration for `Functor Moo'

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Dynamic libraries not of required architecture when installing via cabal (Snow Leopard)
by Malthe Jørgensen 16 Nov '09

16 Nov '09

I'm on Snow Leopard, so I've read numerous posts on the subject of GHC and it's Snow Leopard (non-)compatibility. I've got most things working - I can do a 'cabal update' - something that gave errors before. GHC itself seems to be working fine. I want to install the SDL package from Hackage, which I first tried doing like this: 'cabal install sdl' which gives a bunch of warnings, all looking like this 'ld: warning: in /opt/local/lib/libSDL.dylib, file is not of required architecture' with 'libSDL.dylib' replaced with various *.a and *.dylib files I then found this in a Usenet post about Snow Leopard breaking GHC: -- Once cabal works, options --ld-option=-m32 (and also --gcc-option=- m32) may be used. These options may also be passed to "./Setup configure" And tried this: (I think maybe my problem is that I'm not passing these flags correctly to cabal) 'cabal install sdl --ld-options="-arch i386" --gcc-option=-m32 -- reinstall' Well I get the same errors, well warnings actually. Can I just ignore these warnings? When compiling my small sample code: import Prelude import Graphics.UI.SDL as SDL main = do SDL.init [InitEverything] setVideoMode 640 480 32 [] I again get the warnings: ld: warning: in /opt/local/lib/libSDL.dylib, file is not of required architecture ld: warning: in /opt/local/lib/libSDLmain.a, file is not of required architecture and then a bunch of undefined symbols such as: _SDL_Init _SDL_SetClipRect Well this post basically based on the assumption that the warnings are breaking the SDL library and thereby keeping my code from compiling - so how do I fix those warnings? - Sorry for the long mail Malthe

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index too large
by Nathan M. Holden 13 Nov '09

13 Nov '09

I've been working on a quick program that acts (vaguely) like a console, lets me type notes and outputs them in a .tex file. I figure it's not that complicated, but it uses a vaguely n-ary structure, so that I can have headers and sub-notes, which loos like: data Note = N { nTitle :: {Char}, nBody :: {Char}, nFormat :: Format, subs :: [Note] | E [Char] Format's just a collection of preset formats, so I don't have to dynamically handle that (since I didn't need to for my uses). Anyways, for whatever reason, sometimes while adding notes, it sends me an error: Prelude.(!!): index too large My only 2 uses of that are in getFullName :: [Note] -> [Int] -> [Char] getFullName d [] = "/" getFullName d (x:xs) = "/" ++ getPartName (d!!x) ++ getFullName (subs (d!!x)) xs and moveIn :: [Note] -> [Int] -> [Char] -> Int -> [Int] moveIn [] i m c = reverse i moveIn ((N dt db df du):ds) [] m c = if matchNote (N dt db df du) m then c:[] else moveIn ds [] m (c+1) moveIn n i m c = last i : ((moveIn (subs (n!!(last i)))) (allButLast i) m 0) where the [Int] is the variable I use to track where in the tree I am. Since I'm not moving into a Note that doesn't exist, and the display shouldn't be changing, what could be the problem? This is probably too in-depth a question, now that I think about it...

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maybe this could be improved?
by Michael Mossey 13 Nov '09

13 Nov '09

I've got some code which could be made simpler, I hope. the problem is this: I am implementing a software sampling synthesizer. For a given musical instrument, like piano, there are sound samples in memory. One purchases or creates sample sets. To save time money & resources, most sample sets are not complete---they have samples for only some of the pitches. Perhaps every third pitch has a sample. For the software to produce the sound for a non-included pitch, the software finds the closest included sample and plays it back slightly slower/faster to get the target pitch. That leads to the following code. Any ideas for improvement are welcome. The problem is that there are many cases to check: an empty map? the requested pitch less than all available pitches, greater than all available, or somewhere between? I am specifically writing this to run in O( log n) time. (It would be simpler as O(n).) This particular algorithm probably doesn't need to run in O(log n) time, but I want to do it as an educational experience---I will have other applications that need to use Map in O(log n) time. import Control.Monad.Identity import Control.Monad.Error import Control.Monad import qualified Data.Map as M type Pitch = Int type Sample = String type SampleMap = M.Map Pitch Sample -- Given a SampleMap and a Pitch, find the Pitch in the SampleMap -- which is closest to the supplied Pitch and return that. Also -- handle case of null map by throwing an error. findClosestPitch :: SampleMap -> Pitch -> ErrorT String Identity Pitch findClosestPitch samples inPitch = do when (M.null samples) $ throwError "Was given empty sample table." case M.splitLookup inPitch samples of (_,Just _,_ ) -> return inPitch (m1,_ ,m2) | (M.null m1) && not (M.null m2) -> case1 | not (M.null m1) && (M.null m2) -> case2 | otherwise -> case3 where case1 = return . fst . M.findMin $ m2 case2 = return . fst . M.findMax $ m1 case3 = return $ closest (fst . M.findMax $ m1) (fst . M.findMin $ m2) closest a b = if abs (a - inPitch) < abs (b - inPitch) then a else b

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Re: [Haskell-beginners] Why is type "Integer -> Integer" and not "(Num a) => a -> a"?
by Nathan M. Holden 12 Nov '09

12 Nov '09

This is just a theory, but in my (limited) experience, GHCi is willing to guess at the type of values, where at functions (explicitly or implicitly typed) it can't guess. In your original function, it multiplied a number (Num a) => a by an Int, therefor it must be Int -> Int (because you can't multiply a Double by an Int, don't be crazy. In the new function, though, it can infer the type of the 2, and therefor it can infer it to whatever numerical type you send it. I've run into this problem a few times, but for whatever reason I can't get myself to fall into any of the usual ways I walk blindly into these sorts of problems this time.

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Why is type "Integer -> Integer" and not "(Num a) => a -> a"?
by Dag Hovland 12 Nov '09

12 Nov '09

Hi! I have a problem with understanding some types given by ghc and hugs. The file loaded is: f1 = \x -> x * 2 f2 x = x * 2 After they are loaded I get the following from ghci *Main> :t f1 f1 :: Integer -> Integer *Main> :t f2 f2 :: (Num a) => a -> a *Main> :t \x -> x * 2 \x -> x * 2 :: (Num a) => a -> a I do not understand why f1 is given Integer -> Integer as a type and not the polymorphic (Num a) => a -> a. I believed that f1, f2 and the lambda expression should all have the same type. Similar output to that above is given by Hugs. Thanks, Dag Hovland

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