- Haskell-Cafe - Haskell.org

code generation
by Martin Hofmann 21 Oct '08

21 Oct '08

We try to learn functional programs from examples, but our system is not yet ported to Haskell, though we are working on it. However, we thought about using TH. Do you have any pointers to papers, etc. ? You'll find our project, system and papers here: http://www.cogsys.wiai.uni-bamberg.de/effalip/ Martin

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REMINDER: Haskell Communities and Activities Report
by Janis Voigtlaender 21 Oct '08

21 Oct '08

Dear Haskellers, The deadline for the November 2008 edition of the Haskell Communities and Activities Report is only ten days away. If you haven't already, please write an entry for your new project, or update your old entry. Please mail your entries to hcar(a)haskell.org in plain text or LaTeX format. More information can be found in the original Call for Contributions at: http://www.haskell.org/pipermail/haskell/2008-October/020651.html I look forward to receiving your contributions. Thanks a lot, Janis (current editor) -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:voigt@tcs.inf.tu-dresden.de

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need help making sense of the relative indexing
by Daryoush Mehrtash 21 Oct '08

21 Oct '08

I am trying to make sense of the relative indexing example used in this "Charting Patterns on Price history" paper: http://serv1.ist.psu.edu:8080/viewdoc/download;jsessionid=CC3DEF7277760C535… In Section 3 it defines: type Indicator a = Bar → (Maybe a ) Indicator takes a bar b and returns an indicator value for that bar. .....bar is associated with five basic fields: high, low, open, close price, and transaction volume ...... It is very common while defining indicators to use past values of an indicator. To support this, we have a combina- tor (♯) which enables relative indexing into past data. Given a bar (time), while an indicator, say high , evaluates to the high price at the bar, high ♯ n yields the high price of n th previous bar. (♯) is defined as follows: (♯) :: Indicator a → Integer → Indicator a ind ♯ n = λ b . ind (b − n ) I can't figure out how (b-n) translates to n'th previous bar. Any ideas? daryoush

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external linking traumas
by Chris Dornan 21 Oct '08

21 Oct '08

Hi Folks, I am battling Database.HSQL to get it to work on GHC-6.8.3 running Cygwin/XP and I seem to have found a worthy opponent. I have got the following to work: import Database.HSQL import Database.HSQL.MySQL main :: IO () main = do c <- cnct tbs <- tables c putStr $ unlines $ "Tables:":[" "++tnm|tnm<-tbs] cnct :: IO Connection cnct = connect "host" "database" "user" "password" To work on *ghci* that is, but I can't get it to work with *ghc*. For ghci I use: ghci test.hs -llibmySQL with libmySQL.dll copied into the same directory. If I try the same with ghc: ghc test.hs -package hsql-1.7 -package hsql-mysql-1.7 -L. -llibmysql with libmysql.lib copied into the same directory I get: C:\Program Files\Haskell\hsql-mysql-1.7\ghc-6.8.3/libHShsql-mysql- 1.7.a(MySQL.o)(.text+0x748):fake: undefined reference to `mysql_init@4' C:\Program Files\Haskell\hsql-mysql-1.7\ghc-6.8.3/libHShsql-mysql- 1.7.a(MySQL.o)(.text+0x83f):fake: undefined reference to `mysql_real_connect@32' [...] C:\Program Files\Haskell\hsql-mysql-1.7\ghc-6.8.3/libHShsql-mysql- 1.7.a(MySQL.o)(.text+0x216c):fake: undefined reference to `mysql_stmt_affected_rows@4' collect2: ld returned 1 exit status If I try substituting -llibmySQL (to select the dll) I get the same results. GHC is clearly not resolving the links between Database.HSQL.MySQL and the native libmysql.lib/libmySQL.dll in the way that ghci does with the dll. I must be doing something silly. If someone can tell me I would be grateful. I would also like to know what I need to do to package this up inside CABAL; all my attempts so far have failed. It seems that there has been no maintainer for HSQL for some time, which is a shame, as folks seem to be repeatedly cracking their heads against these problems and re-solving them. I may be able to take on hsql and hsq-mysql myself, but not the other backends. Perhaps it is the prospect of maintaining them all may be stopping folks from stepping forward. (If it is being maintained and I am just making a hash of it then please accept my apologies.) Cheers, Chris

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daemon process
by Rouan van Dalen 21 Oct '08

21 Oct '08

Hi Everyone. I have been learning Haskell the past few months and there are tons of things I still don't understand :) I would like to write a very very simple haskell daemon that disconnects itself from the terminal and sists in the background doing something. What is the easiest way to go about this. I have tried several things but can't seem to get the process disconnected from the terminal. Thanks in advance Rouan Send instant messages to your online friends http://uk.messenger.yahoo.com

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File handles and pipes
by Stephen Hicks 20 Oct '08

20 Oct '08

Hi, I'm trying to understand how to get pipes working in Haskell, in particular with the revamped System.Process (though I've tried similar experiments with System.Posix.IO and Control.Concurrent). Specifically I'm trying to concatenate the output of two system calls into the input of a third. The following code does not get the job done: > import Data.Maybe ( fromJust ) > import System.IO > import System.Process > sink :: String -> IO (Handle,ProcessHandle) > sink s = do let p = (shell s) { std_in = CreatePipe } > (mh, _, _, ph) <- createProcess p > return (fromJust mh,ph) > source :: String -> Handle -> IO () > source s h = do let p = (shell s) { std_out = UseHandle h, > close_fds = False } > (_, _, _, ph) <- createProcess p > waitForProcess ph > return () > main :: IO () > main = do (h,p) <- sink "sort" > source "df" h > source "df -h" h > waitForProcess p > return () When I run this, I see that the filehandle h is closed after the first df, so the output only has one of the two df's included in it. I tried also System.Posix.IO (createPipe, fdToHandle, dup) to make this work, but that gave a "read failed" error. Is there any way to get this done...? In general, given a handful of system calls (and/or read/write handles), is there a way to combine them all together? (I also tried using forkIO on my own function pipe :: [Handle] -> Handle -> IO (), which does what you'd expect given multiple read handles and a single write handle, but couldn't get any output from that at all...) Any help is appreciated, steve

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Re: [Haskell] Probably a trivial thing for people knowing Haskell
by Chris Eidhof 20 Oct '08

20 Oct '08

I think it might be more appropriate to move this discussion to haskell-cafe. On 19 okt 2008, at 17:24, Friedrich wrote: >> Learn to love types: one of the neat things about Haskell is that if >> you can write down the type of a function then you have usually done >> 90% of the work of writing the code for it. > Well I disagree. But that's another story. Well, it's definitely not true when you're starting out with Haskell. The thing is: once you start to think in types it does work like this. You just think: what do I need as my input and what's my output. That's what you write down as your type and you're almost done! It's very similar to test-driven development; the point with TDD is not so much about making sure your program is correct: the big win (for me) is that it helps you think about the design of your program. The same holds for types. > > >> Another is that in >> general, if you can't express the type of a function, it means you >> haven't thought through what you're trying to do. > > No that's not true. The use implies that. However I'm not advice > resistant and will see if I use types. But IMHO that's should be job > of the environment to figure out correctly and most of the time > Haskell does "guess" right. And I surely can ask for the types. I agree. However, sometimes, when things get really complex, you can't figure out a way to write down the code. That's when it can be handy to start out from the types and slowly work towards the definition. At first, you'll think that types are there to make your life harder. After a while, you'll start to love them and to be honest: I feel quite uncomfortable programming in an untyped language these days ;). -chris

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Haskell syntax inside QuasiQuote
by Reiner Pope 20 Oct '08

20 Oct '08

Hi, I have written a toy fixed-length-vector quasiquoter, so that you can write [$vec|1,2|] which has its type inferred as (Vec (S (S Z))) and you can write mkVec :: Double -> Vec (S (S (S Z))) mkVec x = [$vec|1,2,x|] However, these above examples essentially demonstrate the entire syntax it supports: number literals, and anti-quoted variables. I would like to extend my quasiquoter to support Haskell's full expression syntax, so that we can write, for example, mkVec2 x = [$vec| if x > 5 then sin x else cos x |] Is there a simple way to do this, i.e. using existing libraries? What I need is a Haskell expression parser which outputs values of type Language.Haskell.TH.Syntax.QExp, but I can't see one available in the TH libraries, or in the haskell-src(-exts) libraries. Thoughts? Cheers, Reiner

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Is there already an abstraction for this?
by Jeremy Shaw 20 Oct '08

20 Oct '08

Hello, I am trying to figure out if there is an existing abstraction I am missing here. I have an expression data-type: > data Expr > = Quotient Expr Expr > | Product Expr Expr > | Sum Expr Expr > | Difference Expr Expr > | Lit Double > | Var Char > deriving (Eq, Ord, Data, Typeable, Read, Show) > And I want to write a function that will take an expression and automatically apply the identity laws to simplify the expression. The basic identity laws are: a + 0 = a a * 1 = a I can implement these with some 'sugar' as: > identity (Sum (Lit 0) a) = a > identity (Sum a (Lit 0)) = a > identity (Difference a (Lit 0)) = a > identity (Product a (Lit 1)) = a > identity (Product (Lit 1) a) = a > identity (Quotient a (Lit 1)) = a > identity a = a This works fine when the identity only needs to be applied to the root of the expression tree: *Main> ppExpr $ identity (expr "1 + 0") 1 But for more complicated trees it does not fully apply the identity laws: *Main> ppExpr $ identity (expr "0 + (0 + 0) + (0 + 0)") ((0 + 0) + (0 + 0)) What we need to do is first apply the identity function to the children, and then apply them to the parent of the updated children. A first attempt would be to extend the identity function like this: > identity (Sum a b) = identity (Sum (identity a) (identity b)) However, that will not terminate because that same case will keep matching over and over. Another approach is to have two mutually recursive functions like: > identity' (Sum (Lit 0) a) = identityRec a > identity' (Sum a (Lit 0)) = identityRec a > identity' a = a > identityRec (Sum a b) = identity' (Sum (identity' a) (identity' b)) This prevents non-termination, but you have to be careful about calling identity' vs identityRec or you will either re-introduce non-termination, or you may not fully apply the identity laws. Another option to create a helper function like: > -- |Deep map of an expression. > eMap :: (Expr -> Expr) -> Expr -> Expr > eMap f (Sum a b) = f (Sum (eMap f a) (eMap f b)) > eMap f (Difference a b) = f (Difference (eMap f a) (eMap f b)) > eMap f (Product a b) = f (Product (eMap f a) (eMap f b)) > eMap f (Quotient a b) = f (Quotient (eMap f a) (eMap f b)) > eMap f (Var v) = f (Var v) > eMap f (Lit i) = f (Lit i) Now we can easily apply the identity law recursively like: > deepIdentity :: Expr -> Expr > deepIdentity = eMap identity *Main> ppExpr (deepIdentity (expr "0 + (0 + 0) + (0 + 0)")) 0 Sweet! But, having to write eMap out by hand like that somehow feels wrong -- as if I am missing some additional abstraction. In some respects eMap is like a Functor, but not quite. I expect there is a way to implement eMap using Data.Generics, which is perhaps better, but I still feel like that is missing something? Anyway, I just thought I would ask in case someone recognized this pattern and could point me in the right direction. I have attached a working example program you can play with. I would also be interested in alternative approaches besides the ones I outlined. thanks! j. >{-# LANGUAGE DeriveDataTypeable #-} > > import Control.Applicative (Applicative((<*>), pure), (*>), (<*)) > import Control.Monad (ap) > import Data.Generics (Typeable, Data) > import Data.List (isSuffixOf) > import Text.ParserCombinators.Parsec ((<|>)) > import qualified Text.ParserCombinators.Parsec as P > import qualified Text.ParserCombinators.Parsec.Expr as P > import Text.PrettyPrint.HughesPJ ((<+>)) > import qualified Text.PrettyPrint.HughesPJ as H > import Prelude hiding (sum, product) > > data Expr > = Quotient Expr Expr > | Product Expr Expr > | Sum Expr Expr > | Difference Expr Expr > | Lit Double > | Var Char > deriving (Eq, Ord, Data, Typeable, Read, Show) > > instance Applicative (P.GenParser token state) where > pure = return > (<*>) = ap > > parseExpr :: P.GenParser Char st Expr > parseExpr = P.buildExpressionParser optable (lit <|> var <|> parenExpr) > where > parenExpr = > (P.char '(' >> P.skipMany P.space) *> parseExpr <* (P.char ')' >> P.skipMany P.space) > optable = > [ [ P.Infix (P.char '/' >> P.skipMany P.space >> return Quotient) P.AssocLeft ] > , [ P.Infix (P.char '*' >> P.skipMany P.space >> return Product) P.AssocRight ] > , [ P.Infix (P.char '+' >> P.skipMany P.space >> return Sum) P.AssocRight ] > , [ P.Infix (P.char '-' >> P.skipMany P.space >> return Difference) P.AssocLeft ] > ] > lit = > do d <- P.try (P.many1 $ P.oneOf ('-' : '.' : ['0'..'9'])) > P.skipMany P.space > return (Lit (read d)) > var = > do sign <- (P.char '-' >> return (\x -> (Product (Lit (-1)) x))) <|> (return id) > v <- (P.upper <|> P.lower) > P.skipMany P.space > return (sign (Var v)) > > expr :: String -> Expr > expr str = either (error .show) id (P.parse parseExpr str str) > > ppExpr :: Expr -> H.Doc > ppExpr (Lit i) = H.text (let f s = if isSuffixOf ".0" s then init(init s) else s in f $ show i) > ppExpr (Var v) = H.char v > ppExpr (Quotient x y) = H.parens (ppExpr x <+> H.char '/' <+> ppExpr y) > ppExpr (Product x y) = H.parens (ppExpr x <+> H.char '*' <+> ppExpr y) > ppExpr (Sum x y) = H.parens (ppExpr x <+> H.char '+' <+> ppExpr y) > ppExpr (Difference x y) = H.parens (ppExpr x <+> H.char '-' <+> ppExpr y) > -- |Deep map of an expression. > eMap :: (Expr -> Expr) -> Expr -> Expr > eMap f (Sum a b) = f (Sum (eMap f a) (eMap f b)) > eMap f (Difference a b) = f (Difference (eMap f a) (eMap f b)) > eMap f (Product a b) = f (Product (eMap f a) (eMap f b)) > eMap f (Quotient a b) = f (Quotient (eMap f a) (eMap f b)) > eMap f (Var v) = f (Var v) > eMap f (Lit i) = f (Lit i) > identity (Sum (Lit 0) a) = a > identity (Sum a (Lit 0)) = a > identity (Difference a (Lit 0)) = a > identity (Product a (Lit 1)) = a > identity (Product (Lit 1) a) = a > identity (Quotient a (Lit 1)) = a > identity a = a > deepIdentity :: Expr -> Expr > deepIdentity = eMap identity > test :: IO () > test = > do print (ppExpr (deepIdentity (expr "1 + 2"))) > print (ppExpr (deepIdentity (expr "0 + (0 + 0) + (0 + 0)")))

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is 256M RAM insufficient for a 20 million element Int/Int map?
by Thomas Hartman 20 Oct '08

20 Oct '08

{-# LANGUAGE BangPatterns #-} import qualified Data.Map as M import Debug.Trace {- I'm trying to run a HAppS web site with a large amount of data: stress testing happstutorial.com. Well, 20 million records doesn't sound that large by today's standards, but anyway that's my goal for now. I have a standard Data.Map.Map as the base structure for one of my macid data tables (jobs), but I noticed something that is probably causing problems for me. Even a simple 20 million record with int/int key values causes an out of memory error for me in ghci, on a computer with 256M of ram. I'm wondering if there is a data structure that might be more suitable for large recordsets. Or do you just have to use a database, or some sort of file-based serialization, once your records are in the millions? Or is this some weird subtlety of lazy evalution, or some other haskell gotcha? -} size = 2 * 10^7 -- out of memory error t = (M.! size) . myFromList . map (\i->(i,i)) $ [1..size] -- Lists are no problem {- *Main> :! time ghc -e tL testMap.hs (20000000,20000000) 3.38user 0.09system 0:03.53elapsed 98%CPU (0avgtext+0avgdata 0maxresident)k -} tL = (!! (size-1)) . map (\i->(i,i)) $ [1..size] t2 = (M.fromList . map (\i->(i,i)) $ [1..10] ) M.\\ (M.fromList . map (\i->(i,i)) $ [6..15]) -- does this evaluate all of list l, or just whnf? myFromList (!l) = M.fromList l

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