- Haskell-Cafe - Haskell.org

Shortening if-then-else
by Joel Reymont 23 Nov '05

23 Nov '05

Is there a shorter way to write the if-then-else part below? -- tryTakeSeat :: [Word8] -> Word8 -> ScriptState (Maybe Word8) tryTakeSeat _ _ = do ... if (cmdType cmd) /= (CmdSitError Server) then return $ Just seat_num else return Nothing -- Thanks, Joel -- http://wagerlabs.com/

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Re: Frag: a First Person Shooting game
by Joel Reymont 23 Nov '05

23 Nov '05

Mun, I hope this ushers the era of Haskell games! I, for one, was certainly looking for something like this. Did you encounter any difficulties in your development because of Haskell? Was lazy evaluation ever a problem? Why did you choose Yampa to program the game entities? Did this give you a particular advantage? Isn't your frame rate a bit low? Thanks, Joel On Nov 22, 2005, at 6:05 AM, Mun Hon Cheong wrote: > Frag is a 3D First Person Shooting game. > > Features: > > *Yampa, a domain-specific embedded language > for the programming of hybrid systems that > using the concepts of Functional Reactive > Programming (FRP) was used to program the > game entities. -- http://wagerlabs.com/

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re-definition of '.'
by Bulat Ziganshin 23 Nov '05

23 Nov '05

Hello haskell-cafe, if we will define "." as: a.b = b a then we can use "." for 1) field selection 2) sequential functions application in OOP style: [1..100] .map (2*) .sum 3) qualified identifiers - if we treat module name as the record containing all its exported functions: "import M (f,g,h)" defines record M with fields f, g and h The only problem is priority of this operation - for field selection it need to be greater than for application: infixl 11 . while for OOP-styled operations with arguments - its priority must be minimal. one possible solution is rather revolitionary: raise priority of "." if there is no spaces around it, so the following will be interpreted precisely as we want: record.field .List.map (2*) .sum -- Best regards, Bulat mailto:bulatz@HotPOP.com

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Monad Transformer question
by Fan Wu 23 Nov '05

23 Nov '05

Hi Haskell gurus, I'm learning Haskell now and here I'm quite puzzled over some code about Monad Transformers. The code is like: type NDS a = StateT ProblemState [] a getVar :: Var -> NDS (Maybe Value) getVar v = do vs <- gets vars return $ lookup v vs What puzzles me is that, I think the Monad of the do block shall be the "NDS (Maybe Value)" in declaration, but the type of gets is gets :: (MonadState s m) => (s -> a) -> m a So "gets" returns a Monad of type m ([] is this case), which seems to be different from "NDS (Maybe Value)", but GHC does not complain about it. If I comment out the type declaration of "getVar :: Var -> NDS (Maybe Value)" and let GHC interpret the type, then the type of getVar is like: getVar :: (MonadState ProblemState m) => Var -> m (Maybe Value) So does it mean "StateT ProblemState m" and the "m" as in "MonadState ProblemState m" is the same thing? I guess I must missed something when trying to understand the Monad Transformers. Please give me some insights. Thanks, Fan

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RE: [Haskell-cafe] using of data types as kinds
by Simon Peyton-Jones 22 Nov '05

22 Nov '05

Read about Omega! I have not written anything about what I plan to do in GHC, but basically it amounts to allowing you to use a data type as a kind. Busy doing GADTs and impredicativity at the moment though Simon | -----Original Message----- | From: haskell-cafe-bounces(a)haskell.org [mailto:haskell-cafe-bounces@haskell.org] On Behalf Of | Bulat Ziganshin | Sent: 22 November 2005 15:41 | To: haskell-cafe(a)haskell.org | Subject: [Haskell-cafe] using of data types as kinds | | Hello , | | in the HCAR GHC team wrote "On the type system front, we hope to ... | Allow you to use data types as kinds, in a manner similar to Tim | Sheard▓s Omega language". can someone point me where i can read about | this? | | -- | Best regards, | Bulat mailto:bulatz@HotPOP.com | | | | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe(a)haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe

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Existential quantification of environments.
by Keean Schupke 22 Nov '05

22 Nov '05

If I have a function: f x y = add x y and I want to type the function in isolation, then the type of 'add' is essentially carried in the environment... Lets say I want to make this type explicit in the type signature (where f is valid for any a where there is an add function on a - ignoring the class that Haskell would require for the overloading): add :: Int -> Int -> Int add :: Float -> Float -> Float f :: forall a . exists (add :: a -> a -> a) => a -> a -> a or a step further: class Add a where add :: a -> a -> a instance Add Int where ... instance Add Float where ... f :: forall a . Add a => a -> a -> a This seems to suggest: Add a == exists (add :: a -> a -> a) Does this seem in any way right? It seems that the definition of 'f' does require the existance of 'add'. That is the definition is valid iff there exists a function called 'add' of the correct type. Also doesn't the existential quantifier stop you looking inside 'add' - obviously you cannot inspect the definition as it may not be defined (yet?), but presumably you can still apply 'add'. Regards, Keean.

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GHCi always optimizes?
by Dusan Kolar 22 Nov '05

22 Nov '05

Hello all, My question is rather strange, maybe, even if simple: Does ghci always translates with "-O" option set on? I've done some measurements on an application using ghc and ghci. ghc compiled with no opmitization => program running real 988m59.260s user 989m1.325s sys 0m0.704s ghc compiled with optimization set on => program running real 15m54.343s user 15m54.168s sys 0m0.172s ghci alaways: 951.97 secs, 7445117252 bytes (which is 15m51.97sec) To be honest, I don't mind ghci optimizes always but saying it uses the same options as ghc is not entirely true and, for optimization, it's quite painful because one searches error where it is not. (/Understand, I was wondering why program compiled by ghc crashes and/or evaluates so long while running correctly and fast in ghci./) Dusan

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Pickling HList
by Joel Reymont 22 Nov '05

22 Nov '05

Folks, I'm having trouble creating a pickler for HLists and would appreciate a solution. The code for (HCons e HNil) works fine but I get an error trying to implement puHList for (HCons e l) where l is supposed to be (HCons e ...), i.e. another HList. Bar.hs:21:37: Couldn't match the rigid variable e' against PU e' `e' is bound by the instance declaration at Bar.hs:17:0 Expected type: HCons (PU e) l Inferred type: HCons e l In the first argument of puHList', namely l' In the second argument of pair', namely (puHList l)' Failed, modules loaded: none. ---- module Bar where import Data.Word import OOHaskell main = print "We are here!" class HList l => HLPU l where puHList :: HCons (PU e) l -> PU (HCons e l) instance HLPU HNil where puHList (HCons pe HNil) = wrap (\e -> HCons e HNil, \(HCons e HNil) -> e) pe instance HList l => HLPU (HCons e l) where puHList (HCons pe l) = wrap (\(a, b) -> HCons a b, \(HCons a b) -> (a, b)) (pair pe (puHList l)) newtype TourType = TourType TourType_ deriving (Show{-, Typeable-}) newtype AvgPot = AvgPot Word64 deriving (Show{-, Typeable-}) data TourType_ = TourNone | TourSingle | TourMulti | TourHeadsUpMulti deriving (Enum, Show{-, Typeable-}) --- Pickling data PU a = PU { appP :: (a, [Word8]) -> [Word8], appU :: [Word8] -> (a, [Word8]) } pickle :: PU a -> a -> [Word8] pickle p value = appP p (value, []) unpickle :: PU a -> [Word8] -> a unpickle p stream = fst (appU p stream) lift :: a -> PU a lift x = PU snd (\s -> (x, s)) sequ :: (b -> a) -> PU a -> (a -> PU b) -> PU b sequ f pa k = PU (\ (b, s) -> let a = f b pb = k a in appP pb (b, appP pa (a, s))) (\ s -> let (a, s') = appU pa s pb = k a in appU pb s') pair :: PU a -> PU b -> PU (a,b) pair pa pb = sequ fst pa (\ a -> sequ snd pb (\ b -> lift (a,b))) wrap :: (a -> b, b -> a) -> PU a -> PU b wrap (i, j) pa = sequ j pa (lift . i) Thanks, Joel -- http://wagerlabs.com/

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using of data types as kinds
by Bulat Ziganshin 22 Nov '05

22 Nov '05

Hello , in the HCAR GHC team wrote "On the type system front, we hope to ... Allow you to use data types as kinds, in a manner similar to Tim Sheard▓s Omega language". can someone point me where i can read about this? -- Best regards, Bulat mailto:bulatz@HotPOP.com

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How to use notFollowedBy function in Parsec
by Sara Kenedy 22 Nov '05

22 Nov '05

Dear all, Using Parsec, I want to represent a string (of anyToken) not ended with symbol semi (;). I use the command notFollowedby as follows: module Parser where import Parsec import qualified ParsecToken as P import ParsecLanguage langDef::LanguageDef () langDef = emptyDef {reservedOpNames = []} lexer::P.TokenParser() lexer = P.makeTokenParser langDef semi = P.semi lexer str1 :: Parser String str1 = do {str <- many anyToken; notFollowedBy semi; return str} However, when I compile, there is an error. ERROR "Test.hs":17 - Type error in application *** Expression : notFollowedBy semi *** Term : semi *** Type : GenParser Char () String *** Does not match : GenParser [Char] () [Char] I do not know how to fix it. Help me. Thanks for your time.

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