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(no subject)
by u4538637＠anu.edu.au 08 May '08

08 May '08

Hi I have a bit of a dilemma.I have a list of lists, eg, [[1,2,3],[4,5,6],[7,8,9]]. Imagine they represent a grid with 0-2 on the x axis and 0-2 on the y axis, eg, (0,0) is 1, (1,0) is 2, (2,1) is 6, etc and (2,3) is 9. I want to be able to put in the list of lists, and the (x,y) coordinate, and return the value. Also, I need to be able to replace a value in the list. Eg, if I wanted to replace (2,3) with 100, then the output of the expression would be [[1,2,3],[4,5,6],[7,8,100]]. Any help would be great!

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Re: [Haskell-cafe] Random numbers / monads - beginner question
by Thomas Dinsdale-Young 08 May '08

08 May '08

Madoc wrote: Given a list of numbers, I want to modify each of those numbers by adding a random offset. However, each such modified number shall stay within certain bounds, given by the integers minValue and maxValue. After that, I want to continue computation with the resulting list of type [Int]. Personally, I'd do something like this, isolate the IO code outside the algorithm to keep the algorithm pure: modify' :: Int -> Int -> Int modify' offset a = normalize (a + offset) generateInfiniteListOfRandomNumbers :: IO [Int] -- implementation left as an exercise main = do randomNumbers <- generateInfiniteListOfRandomNumbers print $ zipWith modify' randomNumbers [0, 200, 400, 600, 800, 1000] I may be wrong, but generateInfiniteListOfRandomNumbers won't terminate and I think it has to before the next IO action occurs. (Laziness is great, but I don't think you can really do lazy IO like that.) Instead of map :: (a -> b) -> [a] -> [b], I think you are looking for mapM :: Monad m => (a -> m b) -> [a] -> m [b]. * * * *Hope this helps, Thomas >

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Re: [Haskell-cafe] Induction (help!)
by PR Stanley 08 May '08

08 May '08

So, when you apply the function to the first element in the set - e.g. Zero or Nil in the case of lists - you're actually testing to see the function works. Then in the inductive step you base everything on the assumption that p holds for some n and of course if that's true then p must hold for Succ n but you have to prove this by taking Succ from n and thus going bakc to its predecessor which is also the hypothesis p(n). So, to reiterate assumption: if hypothesis then conclusion if p(n) then p(Succ n) proof of assumption if p(Succ n) = Succ(p(n)) then we've won. If pn+1) = p(n) + p(1) then we have liftoff! I'm not going to go any further in case I'm once again on the wrong track. Cheers Paul At 22:43 07/05/2008, you wrote: >On Wed, May 7, 2008 at 9:27 PM, PR Stanley <prstanley(a)ntlworld.com> wrote: > > Hi > > One of you chaps mentioned the Nat data type > > > > data Nat = Zero | Succ Nat > > > > Let's have > > add :: Nat -> Nat -> Nat > > add Zero n = n > > add (Succ m)n = Succ (add m n) > > > > Prove > > add m Zero = m > >To prove this by induction on m, you would need to show: > >1) add Zero Zero = Zero >2) If "add m Zero = m", then "add (Succ m) Zero = Succ m" > >Number (1) is completely trivial, nothing more needs to be said. (2) >is easy, after expanding the definition. > >Here the P I used was P(x) := add m Zero = m, the thing we were trying >to prove. (1) is a base case, P(Zero). (2) is the inductive step, >"If P(m) then P(Succ m)". > >Hoping I don't sound patronizing: if you're still having trouble, then >I suspect you haven't heard what it means to prove an "if-then" >statement. Here's a silly example. > >We want to prove: If y = 10, then y - 10 = 0. > >First we *assume* the condition of the if. We can consider it true. > >Assume y = 10. >Show y - 10 = 0. >Well, y = 10, so that's equivalent to 10 - 10 = 0, which is true. > >Luke

2 1

help in tree folding
by patrik osgnach 08 May '08

08 May '08

Hi. I'm learning haskell but i'm stuck on a generic tree folding exercise. i must write a function of this type treefoldr::(Eq a,Show a)=>(a->b->c)->c->(c->b->b)->b->Tree a->c Tree has type data (Eq a,Show a)=>Tree a=Void | Node a [Tree a] deriving (Eq,Show) as an example treefoldr (:) [] (++) [] (Node '+' [Node '*' [Node 'x' [], Node 'y' []], Node 'z' []]) must return "+∗xyz" any help? (sorry for my bad english)

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runInteractiveProcess and hGetLine on Windows
by Harri Kiiskinen 08 May '08

08 May '08

Hello all, I bumped into a "feature" which might be a bug, but to be certain, I'd like to hear your opinion. I'm running ghc 6.8.2 on Windows XP, and with ghci I do the following: Prelude System.Process System.IO> (inp,outp,err,ph) <- runInteractiveProcess "kpsewhich" ["testfile.txt"] Nothing Nothing ('kpsewhich' is a simple path searching utility used by web2c TeX system, returns the full path to the file if found in the system defined search path, but you all probably know that.) and then: Prelude System.Process System.IO> hGetLine outp which gives me: "./testfile.txt\r" as opposed to "./testfile.txt" which I get on my Linux box. Is the "\r" supposed to be at the end? I thought it is part of the line separator in Windows, and as such, should not be part of the line retrieved? Same thing happens when compiled with ghc. Best Wishes, Harri K.

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How can see old threads.
by Ambrish Bhargava 08 May '08

08 May '08

Hi, I want to know, how can see old threads (may be I can get my answers from there itself)? -- Regards, Ambrish Bhargava

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Re: [Haskell-cafe] Induction (help!)
by PR Stanley 08 May '08

08 May '08

After you grok induction over the naturals, you can start to think about structural induction, which is usually what we use in programming. They are related, and understanding one will help you understand the other (structural induction actually made more sense to me when I was learning, because I started as a programmer and then became a mathematician, so I thought in terms of data structures). Paul: I was hoping that understanding the classic mathematical concept would help me appreciate the structural computer science) variation better. I don't know what it is about induction that I'm not seeing. It's so frustrating! Deduction in spite of the complexity in some parts makes perfect sense. This, however, is a different beast! So let's say you have a tree, and we want to count the number of leaves in the tree. data Tree = Leaf Int | Branch Tree Tree countLeaves :: Tree -> Int countLeaves (Leaf _) = 1 countLeaves (Branch l r) = countLeaves l + countLeaves r We want to prove that countLeaves is correct. So P(x) means "countLeaves x == the number of leaves in x". Paul: By 'correct' presumably you mean sound. First we prove P(Leaf i), since leaves are the trees that have no subtrees. This is analogous to proving P(0) over the naturals. Paul: I'd presume 'proof' here means applying the function to one leaf to see if it returns 1. If I'm not mistaken this is establishing the base case. countLeaves (Leaf i) = 1, by definition of countLeaves. "Leaf i" has exactly one leaf, obviously. So countLeaves (Leaf i) is correct. Now to prove P(Branch l r), we get to assume that P holds for all of its subtrees, namely we get to assume P(l) and P(r). Paul: How did you arrive at this decision? Why can we assume that P holds fr all its subtrees? You can think of this as constructing an algorithm to prove P for any tree, and we have "already proved" it for l and r in our algorithm. Paul: Is this the function definition for countLeaves? This is analogous to proving P(n) => P(n+1) in the case of naturals. Paul:I thought P(n) was the induction hypothesis and P(n+1) was the proof that the formula/property holds for the subsequent element in the sequence if P(n) is true. I don't see how countLeaves l and countLeaves r are analogous to P(n) and P(n+1). So: Assume P(l) and P(r). P(l) means "countLeaves l == the number of leaves in l" P(r) means "countLeaves r == the number of leaves in r" countLeaves (Branch l r) = countLeaves l + countLeaves r, by definition. And that is the number of leaves in "Branch l r", the sum of the number of leaves in its two subtress. Therefore P(Branch l r). Now that we have those two cases, we are done; P holds for any Tree whatsoever. In general you will have to do one such proof for each case of your data structure in order to prove a property for the whole thing. At each case you get to assume the property holds for all substructures. Generally not so many steps are written down. In fact in this example, most people who do this kind of thing a lot would write "straightforward induction", and that would be the whole proof :-) The analogy between structural induction and induction over the naturals is very strong; in fact induction over the naturals is just induction over this data structure: data Nat = Zero | Succ Nat Hope this helps. Luke

7 9

who maintains the POSIX/Unix Unistd.hs?
by Galchin, Vasili 08 May '08

08 May '08

Hello, I am trying to find the definition for the data type "TimeSpec"/"CTimeSpec". thanks, Vasili

2 1

Figuring out if an algebraic type is enumerated through Data.Generics?
by Alfonso Acosta 08 May '08

08 May '08

Hi all, <probably_unnnecesary_background> I'm writing a Hardware-oriented DSL deep-embedded in Haskell (similar to Lava, for those familiar with it). One of the goals of the language is to support polymorphic signals (i.e. we would like to allow signals to carry values of any type, including user-defined types) The embedded compiler has different backends: simulation, VHDL, graphical representation in GraphML ... The simulation backend manages to support polymorphic signals by forcing them to be Typeable. In that way, circuit functions, no matter what signals they process, can be transformed to Dynamic and included in the AST for later simulation. The situation is more complicated for the VHDL backend. The Typeable trick works just fine for translating a limited set of predefined types (e.g. tuples) but not for user-defined types, since the embedded-compiler doesn't have access to the user-defined type declarations. One possible solution to access the type-definition of signal values would be constraining them to instances of Data. Then, we could use dataTypeOf to get access to the type representation. (Another option would be using template Haskell's reify, but I would like to avoid that by now) It would certainly be difficult map any Haskell type to VHDL, so, by now we would be content to map enumerate algebraic types (i.e. algebraic types whose all data constructors have arity zero, e.g. data Colors = Green | Blue | Red) </probably_unnnecesary_background> So, the question is. Is there a way to figure out the arity of data constructors using Data.Generics ? I'm totally new to generics, but (tell me if I'm wrong) it seems that Constr doesn't hold any information about the data-constructor arguments. Why is it so? Do you think there is a workoaround for this problem? (maybe using some other function from Data.Generics different to dataTypeOf?) Thanks in advance, Alfonso Acosta

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Haddock and upload questions?
by Galchin, Vasili 07 May '08

07 May '08

Hello, 1) Can I set up Haddock to run everytime I do a build? 2) http://hackage.haskell.org/packages/upload.html .... - do I have to set up my .cabal in a special way to run "dist"? - once I checked that my Cabal package is "compliant" on this page, how do I get a username and password? Kind regards, Vasili

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license question
by Henrique Ferreiro García 07 May '08

07 May '08

Hello everybody, I have one question regarding a licensing issue. I am developing a library to parse and pretty-print the Core Erlang language and it is heavily based on the modules included in haskell-src. What I want to know is if I have to reproduce all of the LICENSE file included in that package with my name included in the copyright notice or what other thing do I have to do. Thanks in advance. -- Henrique Ferreiro García <hferreiro(a)udc.es>

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Re: [Haskell-cafe] Figuring out if an algebraic type is enumerated through Data.Generics?
by Edward Kmett 07 May '08

07 May '08

On Tue, May 6, 2008 at 12:34 PM, Alfonso Acosta <alfonso.acosta(a)gmail.com> wrote: | So, the question is. Is there a way to figure out the arity of data | constructors using Data.Generics ? | I'm totally new to generics, but (tell me if I'm wrong) it seems that | Constr doesn't hold any information about the data-constructor | arguments. Why is it so? Hmrmm, Playing around with it, I was able to abuse gunfold and the reader comonad to answer the problem : fst $ (gunfold (\(i,_) -> (i+1,undefined)) (\r -> (0,r)) (toConstr "Hello") :: (Int,String)) returns 2, the arity of (:), the outermost constructor in "Hello" A longer version which does not depend on undefined would be to take and define a functor that discarded its contents like: > module Args where > import Data.Generics > newtype Args a = Args { runArgs :: Int } deriving (Read,Show) > tick :: Args (b -> r) -> Args r > tick (Args i) = Args (i + 1) > tock = const (Args 0) > argsInCons = runArgs $ (gunfold tick tock (toConstr "Hello") :: (Args String) Basically all I do is rely on the fact that gunfold takes the 'tick' argument and calls it repeatedly for each argument after a 'tock' base case. The use of the reader comonad or functor is to give gunfold a 'functor-like' argument to meet its type signature. -Edward Kmett

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TimeSpec
by Galchin, Vasili 07 May '08

07 May '08

Hello, I checked Hoogle for the POSIX data type TimeSpec. Nada. Does anybody know where the definition is? Thanks, Vasili

1 0

Still no joy with parsec
by Ross Boylan 07 May '08

07 May '08

I kept getting parse failures when I ran my little Parsec TeX snippet on a sample code. Seeing that ghc6.8 had debugging, I upgraded to it, only to discover that I can't even get the code to compile. $ ghci GHCi, version 6.8.2: http://www.haskell.org/ghc/ :? for help Loading package base ... linking ... done. Prelude> :load g [1 of 1] Compiling Main ( g.hs, interpreted ) g.hs:11:19: Couldn't match expected type `t1 -> GenParser Char () t' against inferred type `CharParser st ()' In the expression: reserved "\\begin" 1 In a 'do' expression: reserved "\\begin" 1 In the expression: do reserved "\\begin" 1 braces (many1 letter) Failed, modules loaded: none. Do parsec and 6.8 just not get along? More generally, how can I go about diagnosing such problems? Since I can't load it, I can't debug it or get :info on the types. It looks as if maybe it's expecting a Monad, but getting a parser. But I don't know why that would have changed vs using 6.6. More questions about the error messages. Where is the expected type, and where is the inferred type, coming from? I'm guessing the expected type is from the function signature and the position inside a do (or perhaps from the argument following the ; in the do?) and the inferred type is what I would just call the type of reserved "begin". And what is the 1 that appears after 'reserved "\\begin"'? An indicator that all occurrences of the text refer to the same spot in the program? Nesting level? Thanks. Ross P.S. There have been some issues with the Debian packaging of ghc6.8, so it's possible I'm bumping into them. I thought/hoped the problems were limited to non i386 architectures. Also, I'm pretty sure that the parsec code used by ghc6.6, ghc6.8, and hugs is all in different files. So conceivably the parsec source differs. I have ghc6 6.8.2-5 and libghc6-parsec-dev 2.1.0.0-2. Source: import Text.ParserCombinators.Parsec import qualified Text.ParserCombinators.Parsec.Token as P import Text.ParserCombinators.Parsec.Language(haskell) reserved = P.reserved haskell braces = P.braces haskell -- TeX example envBegin :: Parser String envBegin = do{ reserved "\\begin" 1 ; braces (many1 letter) }

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Induction (help!)
by PR Stanley 06 May '08

06 May '08

Hi I don't know what it is that I'm not getting where mathematical induction is concerned. This is relevant to Haskell so I wonder if any of you gents could explain in unambiguous terms the concept please. The wikipedia article offers perhaps the least obfuscated definition I've found so far but I would still like more clarity. The idea is to move onto inductive proof in Haskell. First, however, I need to understand the general mathematical concept. Top marks for clarity and explanation of technical terms. Thanks Paul

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ANNOUNCE: category-extras 0.44.2
by Edward Kmett 06 May '08

06 May '08

Dan Doel asked me to roll category-extras into my nascent comonad transformer library, and the result is category-extras 0.44.2! So since Dan's release a couple of weeks ago ( http://www.haskell.org/pipermail/haskell-cafe/2008-April/042240.html) we have added: * Comonad Transformers. Context/ContextT, Reader/ReaderT. * A suite of Bifunctors and combinators in Control.Bifunctor.* -- we attach most of the logic usually associated with a monoidal/comonoidal category to the individual Bifunctor since Hask is a rich category to begin with. * A suite of Functors and combinators in Control.Functor.* * Pointed and copointed functors. * Control.Recursion has been broken out into a Control.Morphism.* and recoded to use a simpler distributive law. * Type Indexed versions of Applicatives, Monads, and Comonads (including Diatchki's Indexed State and Wadler's Delimited Continuation Parameterized Monad) * Parameterized Monads a la Ghani and Johann's paper ICFP 07 paper, and their Applicative and comonadic dual. * Higher-order hylo-, cata- and ana- morphisms. * Higher-order Monads a la Ghani and Johann and their comonadic equivalents. * Kan extensions. * BiKleisli arrows as seen in Uustalu and Vene's Signals and Comonads and SIGFPE's recent posts * The Pointer comonad * Grabbed Iavor Diatchki's value-supply and rolled it in as Control.Comonad.Supply to make it clearer that it is a comonad, and pave the way towards a Supply comonad transformer * A richer set of compositions to allow for construction of comonads and monads not only from adjunctions, but also from pre-composition or post-composition of a monad with a pointed functor, and similarly pre-composition and post-composition of a comonad with a copointed functor. * Generic functor zapping, zipping, unzipping, and cozipping as mentioned recently on http://comonad.com/reader There is still a lot to do in terms of adding back a lot of the documentation from the original, documenting the extensions and fleshing out all of the definable instances as the concepts have grown exceptionally fine-grained. I definitely welcome feedback and additions. In particular if you were using a feature that was supported by the old library or is unnatural to program in the current one, let me know. My goal is to gather a lot of this esoterica into one place and integrate it into something cohesive. HEAD: haddock http://comonad.com/haskell/category-extras/dist/doc/html/category-extras/ darcs http://comonad.com/haskell/category-extras/ Release: package http://hackage.haskell.org/cgi-bin/hackage-scripts/package/category-extras-… -Edward Kmett

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ANNOUNCE: Sessions 2008.5.2
by Matthew Sackman 06 May '08

06 May '08

Howdy, I'm pleased to announce the general availability of Session Types for Haskell, version 2008.5.2. It is available from my website[0], Hackage[1][2] and I've just updated the online tutorial[3] to take into account the recent changes and new features. [0] http://wellquite.org/non-blog/sessions-2008.5.2.tar.gz [1] http://hackage.haskell.org/cgi-bin/hackage-scripts/package/sessions-2008.5.2 [2] http://hackage.haskell.org/ [3] http://wellquite.org/sessions/tutorial_1.html Session types are a means of describing communication between multiple threads, and statically verifying that the communication being performed is safe and conforms to the specification. The library supports multiple, concurrent channels being open and actions upon those channels being interleaved; forking new threads; the communication of process IDs, allowing threads to establish new channels between each other; higher-order channels, allowing an established channel to be sent over a channel to another process; and many other features common to session type systems. The significant new changes in this version are: a) an entirely new means to specify session types. This removes the old absolute indexing, is composeable and far more flexible and powerful than the old system. b) support for higher-order channels and session types allowing channels to be sent and received. This permits additional communication patterns such as two threads which are otherwise unaware of each other being able to communicate with one another by sending and receiving a channel via a common third party - delegation. Tests[4] and an example application[5] are available which should, in combination with the tutorial[6], explain how session types can be used. As ever, any feedback is very gratefully received. [4] http://wellquite.org/non-blog/sessions-browseable/Control/Concurrent/Sessio… [5] http://wellquite.org/non-blog/sessions-browseable/Control/Concurrent/Sessio… [6] http://wellquite.org/sessions/tutorial_1.html Matthew

1 1

Non-Overlapping Patterns
by PR Stanley 06 May '08

06 May '08

Hi isZero :: Int -> Bool isZero 0 = True isZero n | n /= 0 = False The order in which the above equations appear makes no difference to the application of isZero. Does the Haskell interpreter rewrite patterns into one single definition using some sort of switch or if construct? Why does an equation without a guard have to be placed after the more specific cases?To put it another way, why doesn't the interpreter identify the more specific cases and put them before the general ones. Cheers Paul

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Help me speed up my program... or back to the issue of memoization
by Kirill Kuvaldin 06 May '08

06 May '08

Hello, I wrote a program in haskell that works with lattice finite automata (the generalization of notion of finite state automata). Let me post the source code as it is not that long... The problem is that my algorithm that computes the run (see function fun) of an automaton on the given word is not very optimal and takes a loooong time as the input word gets larger... (e.g. try this one "run m3 "11101101110001010101" ") Due to the nature of every haskell function being a referentially transparent, I think I could have speeded up its performance using memoization. I've read the page on haskell wiki (http://www.haskell.org/haskellwiki/Memoization) but it didn't help me because it looks I have to modify the actual function source code to make use of memoized values. What I'm looking for is a kind of a general solution (say, silver bullet :) ) so that I will be able to use my function like > new_run = memoize run and the results of the "new_run" get automatically memoized. Probably it makes sense to memoize deltaext func as well. Is that possible to do that in Haskell?? Thanks a lot! Kirill ======= SOURCE CODE ===== -- data type for lattice data Lattice l = Lattice [l] -- set of lattice elements (l -> l -> l) -- supremum operation (l -> l -> l) -- infimum operation -- returns the lowest lattice element lattice0 (Lattice l s i) = l !! 0 -- returns the greatest lattice element lattice1 (Lattice l s i) = l !! ((length l)-1) -- supremum of 2 lattice elements sup (Lattice l s i) x y = s x y -- infimum of 2 lattice elements inf (Lattice l s i) x y = i x y supremum (Lattice l sup inf) [] = lattice0 (Lattice l sup inf) supremum (Lattice l sup inf) (x:xs) = sup x (supremum (Lattice l sup inf) xs) infimum (Lattice l sup inf) [] = lattice1 (Lattice l sup inf) infimum (Lattice l sup inf) (x:xs) = inf x (infimum (Lattice l sup inf) xs) inf3 (Lattice l s i) x y z = infimum (Lattice l s i) [x,y,z] --- data type for Lattice Automata (LA) data LA l state sym = LA (Lattice l) -- lattice [state] -- set of states [sym] -- alphabet (state -> sym -> state -> l) -- fuzzy transition function (state -> l) -- fuzzy initial state (state -> l) -- fuzzy final state --- extended transition function deltaext :: (Eq state) => (LA l state sym) -> state -> [sym] -> state -> l deltaext (LA l states chars delta sigma0 sigma1) x [] y = if x == y then (lattice1 l) else (lattice0 l) deltaext la@(LA l states chars delta sigma0 sigma1) x (a:w) y = supremum l [ inf l (delta x a z) (deltaext la z w y) | z <- states] -- runs the Lattice Automaton on the given word run la@(LA l states chars delta sigma0 sigma1) w = supremum l [ inf3 l (sigma0 x) (deltaext la x w y) (sigma1 y) | x <- states, y <- states] --- --- examples --- l3 = Lattice [0.0, 0.5, 1.0] max min where max x y = if x > y then x else y min x y = if x < y then x else y m3 = LA l3 ['a', 'b'] ['0', '1'] delta sigma0 sigma1 where delta 'a' '0' 'a' = 1 delta 'a' '0' 'b' = 0.5 delta 'b' '0' 'a' = 0.5 delta 'b' '0' 'b' = 1 delta 'a' '1' 'a' = 0 delta 'a' '1' 'b' = 1 delta 'b' '1' 'a' = 1 delta 'b' '1' 'b' = 1 sigma0 'a' = 1 sigma0 'b' = 0.5 sigma1 'a' = 0.5 sigma1 'b' = 1

3 2

Parsec on TeX
by Ross Boylan 06 May '08

06 May '08

I am new to Haskell and Parsec, and am trying to understand both. I tried to follow the example of how to use Parsec to parse TeX begin/end groups, but can't get it to run. I'm using HUGS -98 on Debian. When I copied the code I got errors about unknown terms (reserved and braces). I've tried to get them from the lexer, but now get this error :load grammar.hsl ERROR "grammar.hsl":21 - Type error in explicitly typed binding *** Term : envEnd *** Type : String -> GenParser Char a [Char] *** Does not match : String -> Parser () Can anyone help me understand what the problem is? Here's the code the caused the above error; I believe the part after --TeX example is verbatim from the Parsec documentation. I picked haskell as the language for to lexer "arbitrarily." import Text.ParserCombinators.Parsec import qualified Text.ParserCombinators.Parsec.Token as P import Text.ParserCombinators.Parsec.Language(haskell) reserved = P.reserved haskell braces = P.braces haskell -- TeX example environment = do{ name <- envBegin ; environment ; envEnd name } <|> return () envBegin :: Parser String envBegin = do{ reserved "\\begin" ; braces (many1 letter) } envEnd :: String -> Parser () envEnd name = do{ reserved "\\end" ; braces (string name) }

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Data structure to manage collection of sets with efficient lookup, intersection?
by Patrick Surry 06 May '08

06 May '08

New to Haskell, with a mental block about how to represent this situation efficiently: I have an unknown function f which is defined on subsets of some universal set (say integers 1...N). I know the values of f for some subsets, and using those can infer values on other subsets. So what I need is a way to manage a collection of subsets s_i (and the associated values f(s_i)) so that I can efficiently (a) check whether a subset s is already 'known' in my collection, and (b) find all subsets t in the collection that intersect s. In a "traditional" language, I'd likely create a dictionary with keys s_i containing the f(s_i) as values, along with a separate dictionary keyed on the elements of the universal set (1...N) in which each entry is a list of all s_i containing the given element of the universal set. Together they let me check, given a subset s, whether I know f(s), and also get the list of all known subsets intersecting s (by the union of the lists of sets containing each element of s). I can't quite wrap my head around how to do this efficiently in Haskell, maybe because of the way the above is duplicating information about the subsets across two different data structures? Any thoughts? Thanks, Patrick ________________________________________________________________________ DISCLAIMER: This e-mail is intended only for the addressee named above. As this e-mail may contain confidential or privileged information, if you are not the named addressee, you are not authorised to retain, read, copy or disseminate this message or any part of it. If you received this email in error, please notify the sender and delete the message from your computer.

3 2

Cabalizing darcs
by Gwern Branwen 06 May '08

06 May '08

So recently I spent a bit of time working on a cabalization of Darcs. It works well for me, and is reasonably easy to apply (attached are three files; do a 'darcs get --lazy http://darcs.net' with Darcs-2 to get the latest, and copy the files into it, the usual autoconf, and it should then work as normal; if this doesn't work for you, I'd appreciate knowing). My question for y'all is: would you find an unofficial version of Darcs - one with the cabalization added and uploaded to Hackage - useful? Or would it just be frustrating and redundant and Not Good? (If yes, any suggestions for a package name? 'darcs' is out, but 'darcs-cb' and 'darcs-cabalized' strike me as horribly clunky or obscure.) -- gwern Threat WANK football Talent anarchy RIT interception Hutsul forschung ISS

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unapplying function definitions?
by PR Stanley 06 May '08

06 May '08

Hi What on earth is unapplying function definitions? The following is taken from chapter 13 of the Hutton book: "...when reasoning about programs, function definitions can be both applied from left to right and unapplied from right to left." Cheers Paul

4 4

Question related to Multi Param Type Classes
by Sai Hemanth K 06 May '08

06 May '08

Hi, I declared a multi param type class as below: *class MyString m c where zIndex :: m -> Int -> c zLength :: m -> Int zPack :: [c] -> m zEquals :: c ->c -> Bool zWrap :: [c] -> (m,AnExistingDataType)* In the end I did not needed it for my program, but for those few minutes I played with the idea, I came across a zillion questions. Can some body help me here please? When I defined a function like below, \begin{code} compareStr::(MyString m c) => (m,Int) ->(m,Int)->Int->Int compareStr (s1,pos1) (s2,pos2) soFar | (pos1 < zLength s1) && (pos2 < zLength s2) = let c1 = zIndex s1 pos1 c2 = zIndex s2 pos2 in if(zEquals c1 c2) then compareStr (s1,(pos1 + 1)) (s2, (pos2 + 1)) (soFar + 1) else soFar | otherwise = soFar \end{code} when I loaded it on ghci ( invoked with -XMultiParamTypeClasses ), I got an essay in greek (or is it latin?), which started something like below: Could not deduce (MyString m c) from the context (MyString m c4) arising from a use of `zLength' at GenericZAlgo.lhs:42:21-31 Possible fix: add (MyString m c) to the context of the type signature for `compareStr' In the second argument of `(<)', namely `zLength s1' In the first argument of `(&&)', namely `(pos1 < zLength s1)' In a pattern guard for the definition of `compareStr': (pos1 < zLength s1) && (pos2 < zLength s2) And if I let the type inference run the show, the type it shows is : compareStr :: (MyString t c2, MyString m c2, Num a, MyString t1 c2, MyString t c, MyString t1 c1) => (t, Int) -> (t1, Int) -> a -> a And puzzlingly, if I try the same thing myself and reload it on the ghci,I still get similar error. I can go on about other 'puzzles'. Can some one please tell me whats going on here? Apologies, if it is something that's been documented left to right, top to bottom, but I could not get anywhere readily. Any pointers to the right documentation will be much appreciated. On the surface it looks like something that n00bs like yours truly better stay away. But I have a feeling that this one may throw some more light on type inference. Many thanks

5 4

Grokking monads by not overlooking concatMap
by Achim Schneider 06 May '08

06 May '08

guard True = return () guard False = [] -- the textbook example f = do x <- [1..4] y <- [2..8] guard (x*y==8) return (x,y) -- the translation. f' = [1..4] >>= (\x -> [2..8] >>= (\y -> (if (x*y == 8) then return () else []) >> return (x,y))) -- another one. note that we can move the final -- return out of the parens. It would even work without it, but we'd -- get a list containing many empty lists: concat fixes that. f'' = [1..4] >>= (\x -> [2..8] >>= (\y -> if (x*y == 8) then [(x,y)] else [] >>= return )) -- bind desugared. Actually, it seems to be impossible to -- write this in monadic notation and I'm getting quite confused. f''' = concatMap (\x -> concatMap (\y -> if x*y == 8 then [(x,y)] else []) [2..8]) [1..4] -- That's the one I've been looking for. Remember that -- return e = [e] f'''' = concatMap (\x -> concatMap (\y -> concatMap (\_ -> [(x,y)]) (if x*y == 8 then [()] else [])) [2..8]) [1..4] Morale: Premature term elimination is the root of all misunderstanding. Haskell wins the wickedness of design contest by using [()] and [] as truth values. -- (c) this sig last receiving data processing entity. Inspect headers for past copyright information. All rights reserved. Unauthorised copying, hiring, renting, public performance and/or broadcasting of this signature prohibited.

2 5

A common pattern
by Andrew Coppin 05 May '08

05 May '08

I've found myself writing code like this several times now. Is there a better way? read_args h = do line <- hGetLine h case line of "." -> return [] ('#':y) -> do ys <- read_args h return (y:ys)

4 5

ANN: ReviewBoard 0.1 bindings
by Adam Smyczek 05 May '08

05 May '08

This package is part of a development tool designed to monitor code changes, analyze dependencies etc. Actually, we are still in process to develop the tool and this binding is the first functional ready package others might be interested in as well. The package contains a basic set of API calls to ReviewBoard server and a small demo app, a command line tool that makes submitting new code review requests as easy as: svn diff | mkrr -r [reviewers] (ReviewBoard does not support darcs) :( For details see haddock documentation. The ReviewBoard project page: http://code.google.com/p/reviewboard/ And the bindings on hackage: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/ ReviewBoard-0.1 A thanks to this list for help with this project and help in getting started with Haskell by following many good threads. Adam

2 1

Ann: Win32-notify 0.1
by Niklas Broberg 05 May '08

05 May '08

Hi all, I wanted to use hinotify, which gives notifications on file system changes, but found (quite naturally) that it wouldn't run on my windows machine (since it's a binding to the linux kernel). So I started writing a library that would give similar functionality on Windows. This is the first release of that, more a proof-of-concept than anything useful at this stage: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/Win32-notify-0.1 Still, feel free to test it, and I would love feedback on everything from desired API to suggestions on how to implement the low-level bindings more efficiently. Cheers, /Niklas

6 6

sound synthesis
by Thomas Girod 05 May '08

05 May '08

Hi there. Following this advice (http://reddit.com/info/6hknz/comments/c03vdc7) I'm posting here. Recently, I read a few articles about Haskell (and FP in general) and music/sound. I remember an article ranting about how lazy evaluation would be great to do signal processing, but it was lacking real world example. I tried to do a little something about it, even though I'm still an haskell apprentice. So, here I come with a small bit of code, waiting for your insights to improve it. The task is to generate a sine wave and pipe it to /dev/dsp on my linux box. There is probably a nicer way to make some noise, like using SDL audio API bindings, but I didn't take the time to poke around this yet. So here it is : > module Main where > import qualified Data.ByteString as B > import Data.Word > import IO (stdout) > rate = 44100 > sinusFloat :: [Float] > sinusFloat = map (\t -> (1 + sin (t*880*2*pi/rate)) / 2) [0..44099] > sinusWord :: [Word8] > sinusWord = map (\s -> floor (s * max)) sinusFloat > where max = 255 > byte = B.pack sinusWord > main = B.hPut stdout byte It is supposed to generate a 880hz sine wav playing for one second, by typing ./bin > /dev/dsp, assuming your soundcard has a 44100hz samplingrate. /dev/dsp is supposed to receive its audio flux as an unsigned byte stream, that's why I'm converting my sine from [-1;1] to [0;1] and then to [0;255] Word8. However, I must miss something because the sound does not have the right frequency and is played too long. I guess the default sound format is 44100hz 16bits stereo, which would explain why it doesn't behave as expected. I'm wondering how I could convert a [Word16] to ByteString, and how I could use lazy evaluation to generate an infinite sine that stops with the interupt. Thomas

6 5

Re: Understanding tail recursion and trees
by Edsko de Vries 04 May '08

04 May '08

Hi, Thanks to Miguel for pointing out my silly error. So at least my understanding of tail recursion is correct :) So then the question becomes: what *is* the best way to write this function? One version I can think of is > ecount :: [Tree] -> Integer -> Integer > ecount [] acc = acc > ecount (Leaf _ : ts) acc = ecount ts $! (acc + 1) > ecount (Branch t1 t2 : ts) acc = ecount (t1 : t2 : ts) acc which essentially maintains an explicit stack and runs on all trees. Are there better ways to do this? Thanks again and sorry for my mistake, Edsko

5 7

castIOUArray and hPutArray redux
by Ben 03 May '08

03 May '08

hi haskellers, have the issues with castIOUArray (and thus hGetArray, hPutArray) in Data.Array.IO discussed below been resolved? http://www.haskell.org/pipermail/libraries/2003-January/thread.html here is a (trivial) program which has rather unexpected behavior. (i'm switching to Data.Binary in the meantime.) import Data.Array.IO import System.IO (IOMode(..), openBinaryFile, hClose) dumpArray arr name = do h <- openBinaryFile name WriteMode w <- castIOUArray arr (l,u) <- getBounds w let len = u-1+1 hPutArray h w len hClose h return len loadArray name num = do arr <- newArray_ (1, num) h <- openBinaryFile name ReadMode read <- hGetArray h arr num if not(read == num) then error "Incorrect number of bytes read in from array!" else return arr len = 20 main = do arr <- newArray (1, len) 1::IO(IOUArray Int Int) size <- dumpArray arr "foo" arr2' <- loadArray "foo" size arr2 <- (castIOUArray arr2')::IO(IOUArray Int Int) e1 <- getElems arr e2 <- getElems arr2 print e1 print e2 print size print $ show (arr == arr2) best, b

3 3

Understanding tail recursion and trees
by Edsko de Vries 03 May '08

03 May '08

Hi, I am writing a simple compiler for a small DSL embedded in Haskell, and am struggling to identify and remove the source of a stack error when compiling larger examples. To understand the issues better, I was playing around with tail recursion on trees when I came across the following problem. Suppose we want to count the number of leaves in a tree. The obvious naive (non tail-recursive) will run out of stack space quickly on larger trees. To test this, I defined a function that generates left (gentreeL, code below) or right (gentreeR) biased trees, that look like * * / \ / \ * * * * / \ / \ * * * * . . . . n n respectively; that is, a tree of depth n, with on the right (or the left) leaves only). Now, we can define define two traversals: one that has a tail call for the left subtree, after having traversed the right (acountL, below); and one that has a tail call for the right subtree, after having traversed the left (acountR). I was expecting acountL to work on the left biased tree but not on the right biased tree -- and that assumption turned out to be correct. However, I was also expecting (by "duality" :) acountR to work on the right biased tree, but not on the left biased tree, whereas in fact it works on both! (Indeed, it works on reallybigtree3, which combines the left and right biased trees, as well.) Can anyone explain why this is happening? Why is acountR not running out of stack space on the left biased tree? Also, if this is quirk rather than something I can rely on, is there a way to write a function that can count the number of leaves in reallybigtree3 without running out of stack space? Thanks (code follows), Edsko > data Tree = Leaf Integer | Branch Tree Tree > -- naive count > ncount :: Tree -> Integer > ncount (Leaf _) = 1 > ncount (Branch t1 t2) = ncount t1 + ncount t2 > -- generate left-biased tree (right nodes are always single leaves) > gentreeL :: Integer -> Tree > gentreeL 0 = Leaf 0 > gentreeL n = Branch (gentreeL (n-1)) (Leaf 0) > > -- generate right-based tree (left nodes are always single leaves) > gentreeR :: Integer -> Tree > gentreeR 0 = Leaf 0 > gentreeR n = Branch (Leaf 0) (gentreeR (n-1)) > > -- test examples > reallybigtree1 = gentreeL 2000000 > reallybigtree2 = gentreeR 2000000 > reallybigtree3 = Branch (gentreeL 2000000) (gentreeR 2000000) > > -- count with tail calls for the left subtree > acountL :: Tree -> Integer -> Integer > acountL (Leaf _) acc = acc + 1 > acountL (Branch t1 t2) acc = acountL t1 $! (acountL t2 acc) > > -- count with tail calls for the right subtree > acountR :: Tree -> Integer -> Integer > acountR (Leaf _) acc = acc + 1 > acountR (Branch t1 t2) acc = acountR t2 $! (acountL t1 acc)

3 2

Re: Function Type Calculation (Take 2)
by PR Stanley 02 May '08

02 May '08

Just in case anyone missed this: [1] funk f x = f (funk f) x f :: a x :: b funk f x :: c therefore funk :: a -> b -> c RHS f (funk f) x :: c f (funk f) :: d -> c x :: d f :: e -> d -> c funk :: h -> e f :: h unification f :: a = h = (e -> d -> c) x b = d No. x :: b = d (a typo?) Paul: What's wrong with x being of type b and of type d? Could you perhaps explain the error please? Don't forget also that funk :: a -> b -> c = h -> e, which means that e = b -> c Paul: is that something to do with partial application? (funk f) is a partially applied function, correct? Again an explanation would be appreciated. therefore funk :: ((h -> e) -> b -> c) -> b -> c No. I don't understand where you've got this expression from. It's funk :: a -> b -> c = (e -> d -> c) -> b -> c = ((b -> c) -> b -> c) - > b -> c According to GHCi: Prelude> let funk f x = f (funk f) x Prelude> :t funk funk :: ((t1 -> t) -> t1 -> t) -> t1 -> t which is about the same. Thanks Paul

2 1

elementary Maybe Monad problem .. sigh
by Galchin, Vasili 02 May '08

02 May '08

data Bozo = Bozo { id :: Int } bonzo :: Maybe Bozo -> IO () bonzo maybe_bozo = do if maybe_bozo == (Just (Bozo x)) then return () else return () ~ I want "x" to be a pattern that matches "id" .... ?? Kind regards, Vasili

5 6

Writing an 'expect'-like program with runInteractiveCommand
by Graham Fawcett 02 May '08

02 May '08

Hi folks, I would like to communicate with an external, line-oriented process, which takes a sequence of one-line commands, each returning an arbitrary number of lines, and waits for another command after each response. So, something like: sendCmd :: (Handle, Handle) -> String -> IO [String] ... main = do handles <- connectToExternalProcess sendCmd handles "do something" resp <- sendCmd "get results" -- needs strict I/O, before "quit"? sendCmd "quit" mapM_ putStrLn resp I've tried using runInteractiveCommand, and several combinations of hFlush, hWaitForInput, etc., but I can't find a combination that actually works. I know this is a sketchy description, but can anyone offer some sample code, or point me toward a program that has similar behaviour? Thanks, Graham

3 3

Test.QuickCheck.Gen
by Tony Morris 02 May '08

02 May '08

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 I am trying to understand the QuickCheck source in a bit of depth and I have noted that a Gen x is a function Int -> StdGen -> x, however, some of the combinators can at times, fail to ever produce a result by throwing an error. I include an example using ghci: *Test.QuickCheck> (f 0 (mkStdGen 0)) :: Int *** Exception: Prelude.(!!): negative index It seems this can be alleviated by changing Gen x to Int -> StdGen -> Maybe x and having the generator produce Nothing when a combinator fails. However, now the function promote cannot be written (I have tried and run into the conundrum of writing the function (a -> Maybe b) -> Maybe (a -> b)), which is used to generate functions. So, I'm caught on the fence about whether there is a possible improvement here (for example, should the elements function throw the error?), or if the current scenario is acceptable. I'm seeking comments about this, thanks! - -- Tony Morris http://tmorris.net/ -----BEGIN PGP SIGNATURE----- Version: GnuPG v1.4.6 (GNU/Linux) Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org iD8DBQFIGjO6mnpgrYe6r60RAtohAKCBsl1lHxuNrBaLHuqwCN58PBHCIACbBALk YWkBkw9o9NUQbr+lgo0rXE0= =JQlH -----END PGP SIGNATURE-----

1 0

ANN: Haddock version 2.1.0
by David Waern 02 May '08

02 May '08

Hi everyone, I'm pleased to announce Haddock 2.1.0. Hackage page: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/haddock-2.1.0 Changes since last version: * Fix a bug that made links point to the defining module instead of the "best" one (e.g Int pointing to GHC.Base instead of Data.Int) * Fix a couple of smaller bugs * The representation of DocName was changed in the library * Add a flag --no-warnings for turning off warnings David

2 2

Haskell Web server
by Cetin Sert 01 May '08

01 May '08

Hi, What is the fastest way to set up a web server that can redirect requests to a haskell application and serve results returned by it? I need to demonstrate a simple visualization tool I have written for analytic tableaux on Monday and need something easy and simple. Best Regards, Cetin Sert

3 2

approximating pi
by ry dahl 01 May '08

01 May '08

By picking points randomly from a square one can calculate pi. Keep track of how many points from the square you pick lay in the inscribed circle. Supposing the square's edges are length 2, then the inscribed circle has radius 1. Thus the area of the circle is pi*r^2 = pi. The area of the square is 4. The ratio of points picked from the circle to the total number of picked points will converge to pi / 4. What is the best way to express this algorithm in Haskell? ry

7 9

[help] Program immplementing an alarm
by Antonio Regidor García 01 May '08

01 May '08

Hi, I'm trying to implement an alarm in Haskell and wrote the following code: http://hpaste.org/7201 But it doesn't work as expected. The expected behaviour is: The user enters a time in minutes. If the time is reached, a bell rings. If, before the time is reached, he press enter, the program finish and prints the elapsed time. The actual behaviour is: The user enters a time in minutes. If time is reached, nothing happens. If the user press enter, the elapsed time is printed and the program finish. Also, if he pressed enter after the time had elapsed, the bell rings. Some ideas? Antonio Regidor García ______________________________________________ Enviado desde Correo Yahoo! La bandeja de entrada más inteligente.

2 2

ANN: Graphics.Rendering.Diagrams 0.1
by Brent Yorgey 01 May '08

01 May '08

Hi all! I am very pleased to announce the release of Graphics.Rendering.Diagrams, an embedded domain-specific language for creating simple pictures and diagrams, built on top of the Cairo vector graphics library. It's still somewhat in its infancy, but already beginning to be useful and I wanted to get something released as soon as possible. More info and examples here: http://byorgey.wordpress.com/2008/04/30/ new-haskell-diagrams-library/ Download the package here: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/diagrams-0.1 If it looks interesting to you, please give it a try, report bugs, request features, and so on. If it looks too simplistic to be of use to you -- just wait for a future version! I've got tons of ideas for features. If you're interested in contributing, I'll have a darcs repo up (most likely at http://code.haskell.org/diagrams/) as soon as someone creates a directory for me on code.haskell.org. Enjoy! -Brent

1 1

question about GHC and Unicode
by Daniel Gee 01 May '08

01 May '08

In GHC there's a GHC.Unicode library, but for a string such as "ΧΑΟΣΣ", a GHC compiled program prints it as a string of unknown characters, and in the interpreter, the string evaluates to a string of escape sequences instead of displaying properly. Is there a way to get/activate unicode support in GHC?

9 13

Bug or feature with -XFlexibleContects?
by John Goerzen 01 May '08

01 May '08

Hi, I noticed something odd yesterday. I had a function of this type: divByGeneric :: (Integral a, MonadError (DivByError a) m) => a -> [a] -> m [a] If I left off the type signature, ghci (with no command-line parameters) would load the file just fine, and :t would give this same type back. But if I put the type signature in there -- even an exact cut and paste from ghci's :t -- it would complain that it needs -XFlexibleContexts. It seems odd to me that ghci can infer a type that is illegal for the way it was run. Thoughts? This is GHC 6.8.2, BTW. -- John

2 2

Re: [Haskell-cafe] HaXml and the XHTML 1.0 Strict DTD
by Marc Weber 30 Apr '08

30 Apr '08

Also have a look at the HaXml page. A branch is listed there passing more tests if I recall correctly. Don't know wether those changes have been merged back yet. I haven't tried it. Marc Weber

1 0

Function Type Calculation (Take 2)
by PR Stanley 30 Apr '08

30 Apr '08

In case you missed it the first time here is my query again: Hi I know we've already looked at the topic of function type calculation though last time I didn't have the chance to go through it thoroughly. So here it is again. Apologies for the repetition. I've had a try at calculating function types for two examples below. So to start with I'd be grateful for an assessment of my efforts. All comments are welcome. Thanks, Paul [1] funk f x = f (funk f) x f :: a x :: b funk f x :: c therefore funk :: a -> b -> c RHS f (funk f) x :: c f (funk f) :: d -> c x :: d f :: e -> d -> c funk :: h -> e f :: h unification f :: a = h = (e -> d -> c) x b = d therefore funk :: ((h -> e) -> b -> c) -> b -> c [2] w f = f f Assigning Types f :: a w f :: b therefore w :: a -> b RHS f f :: b f :: c -> b f :: c f :: a = b = c = (c -> b) therefore w :: (a -> a) -> a

2 2

Re: [Haskell] Matrix multiplication
by Don Stewart 30 Apr '08

30 Apr '08

> droundy > On Wed, Apr 23, 2008 at 05:24:20PM +0100, Sebastian Sylvan wrote: > > On Wed, Apr 23, 2008 at 5:01 PM, Tillmann Vogt <Tillmann.Vogt <at> rwth-aachen.de> > > wrote: > > > > > Hi, > > > > > > I am currently experimenting with parallelizing C-programs. I have > > > therefore written a matrix vector multiplication example that needs 13 > > > seconds to run (5 seconds with OpenMP). Because I like Haskell I did the > > > same in this language, but it takes about 134 seconds. Why is it so slow? > > > Does someone have an idea? > > > > > Yes, in the C version you use unboxed arrays, in the Haskell version you use > > a linked list of linked lists. Naturally the latter will take up more space, > > require more work to index, and will thrash the cache quite a bit. > > In fact, I'm impressed that Haskell can come within a factor of 10, given > that it's using such a challenging data type! (I wonder if this may be due > to inefficiencies in the C code, although I haven't looked at it.) I had a look at this code, using the (unboxed, strict, fused) data parallel arrays library (http://darcs.haskell.org/packages/ndp) We get rather nice code. The original C program (modified, to sum the result, rather than just filling the array and throwing it out): #include <stdio.h> #include <stdlib.h> #include <time.h> #define M 4000 #define N 4000 #define IT 100 double a[M], b[M][N], c[N]; int main(int argc, char *argv[]) { double d; double sum; int i, j, l; time_t start,end; printf("Initializing matrix B and vector C\n"); for(j=0; j<N; j++) c[j] = 2.0; for(i=0; i<M; i++) for(j=0; j<N; j++) b[i][j] = j; printf("Executing %d matrix mult. for M = %d N = %d\n",IT,M,N); time (&start); sum = 0; for(i=0; i<M; i++) for (j=0; j<N; j++) sum += b[i][j]*c[j]; time (&end); printf ("result: %.2lf\n", sum ); d = difftime (end,start); printf ("calculation time: %.2lf seconds\n", d ); return 0; } And then rewritten in a data parallel style (but note, no parallelism here, just using the nice interface to STUArrays, with stream fusion): import Data.Array.Parallel.Unlifted import Text.Printf m = 4000 n = 4000 it = 100 main = printf "result: %2f\n" . sumU . zipWithU (*) (concatSU b) $ repeatU m c where -- c[N] c = replicateU n (2.0 :: Double) -- b[M][N] b = replicateCU m t where t = mapU fromIntegral (enumFromToU 0 (n-1)) :: UArr Double Finds 4 fusion sites, and runs in: $ time ./a result: 63984000000.0 ./a 0.10s user 0.00s system 112% cpu 0.089 total While the C program, $ time ./a.out Initializing matrix B and vector C Executing 100 matrix mult. for M = 4000 N = 4000 result: 63984000000.00 calculation time: 0.00 seconds ./a.out 0.14s user 0.09s system 92% cpu 0.248 total GHC rocks :) NDP programming encourages a nice combinator style too, which is interesting (and Haskell-ish). The compiler is able to fuse array most of the arrays, (and Roman reports this should do rather better with ghc head). Anyway, we've a few new (fast!) array options on the horizon, looking promising. For more info, check out the data parallel arrays page, http://haskell.org/haskellwiki/GHC/Data_Parallel_Haskell -- Don

1 0

Caching the Result of a Transaction?
by Jake Mcarthur 30 Apr '08

30 Apr '08

I have a problem I've been trying to work around using the existing STM API, and so far it seems that I may be unable to do it. For more background, see my blog post at <http://geekrant.wordpress.com/2008/04/25/stm-caching-need/ >. Here, for brevity, I will only describe exactly what I think I need, not what it's for. Say I have a function f :: STM a. The transaction reads from one or more TMVars, performs some computation, and returns the result in the STM monad. Also, in this scenario, it is known that once the TMVars have values, those values will never be changed again (write once, read many, somewhat like IVars before they were removed). Now say I try to use this function as so. liftM2 (,) f f So the desired result is a pair in the STM monad where both components are the result from f. The problem I have is that, in the above example, the TMVars are all read twice and the computations are all performed twice, once for each of the components of the resulting pair. In many cases, this may be the correct thing to do because the values of the TMVars may have changed, but what about this case where I _know_ that the values have not been modified? What I need is a way to cache the result of f so that future uses of f don't have to reread from the TMVars, even across multiple transactions, maybe even leading to the eventual garbage collection of the TMVars if they are not used elsewhere. Right now I think the only way to do this would be to change the STM implementation slightly and create a new primitive function. If there is a way to do something like this with the current STM API, I would love to hear suggestions. Any ideas? - Jake McArthur

5 14

Re: [Haskell-cafe] A bright future for Haskell
by Don Stewart 30 Apr '08

30 Apr '08

ajb: > G'day all. > > Quoting John Peterson <jpeterson(a)western.edu>: > > >Especially if SPJ decides to grow a beard. Unfortunately Paul is > >now clean shaven so maybe Haskell is in trouble. > > This explains why Clean never made it: Rinus Plasmeijer can't compete > with Phil Wadler in the beard department. > > I should point out, for fairness, tht Mark Jones has suitable facial > hair, and John Hughes and Ralf Hinze are not fair behind. John Peterson > clearly should complete his beard, though. And in the younger generation, Ian Lynagh is surely critical to Haskell's on going success... -- Don

1 0

A bright future for Haskell
by John Peterson 30 Apr '08

30 Apr '08

Especially if SPJ decides to grow a beard. Unfortunately Paul is now clean shaven so maybe Haskell is in trouble. http://blogs.microsoft.co.il/blogs/tamir/archive/2008/04/28/computer-langua… John

2 1

catching fails on read
by Jacques Bergeron 30 Apr '08

30 Apr '08

I'm new to Haskell and have a stupid question. How can a catch a parsing error on a read? Something like let s = "e" in read s :: Int I've been searching for a trick and couldn't find one. I want to enter commands directly from the keyboard and don't want the program to abort if I do a typing error. Regards, Jacques

2 2