February 2009 - Haskell-Cafe

Iteratee-based IO and lightweight monadic regions in the wild
by Artyom Shalkhakov 14 Feb '09

14 Feb '09

Hello, Is anybody planning to use these shiny new ways for doing IO? I'm also interested in a fair comparison of ByteString/Binary with iteratee-based IO. Has anybody done this? Regards, Artyom Shalkhakov.

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Re: [Haskell-beginners] Re: permuting a list
by Daniel Fischer 14 Feb '09

14 Feb '09

Am Samstag, 14. Februar 2009 16:37 schrieb Heinrich Apfelmus: > > That of course begs the question whether there is a faster but purely No, it didn't beg any question. (Sorry for being a humourless pedant here) > functional algorithm for generating random permutations without indexes > and arrays? > > The answer is a resounding "yes" and the main idea is that shuffling a > list is *essentially the same* as sorting a list; the minor difference > being that the former chooses a permutation at random while the latter > chooses a very particular permutation, namely the one that sorts the input. > > For the full exposition, see > > http://apfelmus.nfshost.com/random-permutations.html Excellent work, thanks. > > > Regards, > apfelmus Cheers, Daniel

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Possible bug?
by Peter Verswyvelen 14 Feb '09

14 Feb '09

Could it be considered a bug when a function compiles fine without type signature, but when you add the type signature that GHCi reports with :type, it fails to compile? I am using functional dependencies, and I know these are not the best way to do it. Can all functional dependencies be completely replaced with associated types when using GHC 6.10.1? I tried once but I failed.

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space leak with 'concat' ?
by Henning Thielemann 14 Feb '09

14 Feb '09

$ ghc +RTS -M16m -c30 -RTS -e 'concat $ repeat "bla"' This breaks down after a while, also if I increase the memory restriction: ... ablablablablablablablablablablablablablablablablablablablablablaHeap exhausted; Current maximum heap size is 15998976 bytes (15 Mb); use `+RTS -M<size>' to increase it. Whereas this one works: $ ghc +RTS -M16m -c30 -RTS -e 'cycle "bla"' 'concat' seems to be the culprit. What's so bad about it?

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IO semantics and evaluation - summary
by Gregg Reynolds 14 Feb '09

14 Feb '09

Many thanks to everybody who tried to set me straight on the thread about IO monad and evaluation semantics. I've begun summarizing the info, and I believe I've come up with a much better way of explaining IO; just flip the semantic perspective, and think in terms of interpretations instead of actions. Voila! Oxymoron (values that perform actions) eliminated. See the "Computation considered harmful" and "Fixing Haskell IO" articles at http://syntax.wikidot.com/blog Naturally I would be grateful for any corrections/comments. Thanks, gregg

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Class Instances
by Cetin Sert 14 Feb '09

14 Feb '09

Hi, class Processor a where ready :: (forall b c. a → [b → c]) {- instance Processor (b → c) where ready = repeat instance Processor [b → c] where ready = id-} doSth :: (Show p, Processor p) ⇒ p → IO () doSth p = print p ------------------------------- Why can I not declare the above instances and always get: message.hs:229:10: Couldn't match expected type `b' against inferred type `b1' `b' is a rigid type variable bound by the instance declaration at message.hs:228:20 `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat message.hs:229:10: Couldn't match expected type `c' against inferred type `c1' `c' is a rigid type variable bound by the instance declaration at message.hs:228:24 `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 Expected type: b -> c Inferred type: b1 -> c1 In the expression: repeat In the definition of `ready': ready = repeat message.hs:232:10: Couldn't match expected type `b1' against inferred type `b' `b1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:19 `b' is a rigid type variable bound by the instance declaration at message.hs:231:20 Expected type: [b1 -> c] Inferred type: [b -> c1] In the expression: id In the definition of `ready': ready = id message.hs:232:10: Couldn't match expected type `c1' against inferred type `c' `c1' is a rigid type variable bound by the type signature for `ready' at message.hs:226:21 `c' is a rigid type variable bound by the instance declaration at message.hs:231:24 Expected type: [b -> c1] Inferred type: [b1 -> c] In the expression: id In the definition of `ready': ready = id Is there a way around this? Regards, CS

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Another point-free question (>>=, join, ap)
by Edsko de Vries 14 Feb '09

14 Feb '09

Hi, I can desugar do x' <- x f x' as x >>= \x -> f x' which is clearly the same as x >>= f However, now consider do x' <- x y' <- y f x' y' desugared, this is x >>= \x -> y >>= \y' -> f x' y' I can simplify the second half to x >>= \x -> y >>= f x' but now we are stuck. I feel it should be possible to write something like x ... y ... f or perhaps f ... x ... y the best I could come up with was join $ return f `ap` x `ap` y which is not terrible but quite as easy as I feel this should be. Any hints? Edsko

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Language popularity
by Henk-Jan van Tuyl 13 Feb '09

13 Feb '09

Yesterday I saw Haskell mentioned for the first time in a magazine, Bits & Chips. It is a magazine for professionals, about hardware and software; the article was about the domain specific language Cryptol from Galois. In the same issue of the magazine, there was an article saying that the company Tiobe has proclamed C the language of the year 2008, because of it's growth in popularity of almost two percent. The current statistics can befoud at [1]. The most popular functional language at the moment is LOGO [2], at the 15th place in the top 50 (from 22nd place a year ago). Haskell is at the 35th place (no indication of the score last year). A quick search in the Web Archive [3] reveals that Haskell was at the 41st place in the index in June 2007. My own research, using Google: Search Hits ------------------------------- Java programming 20.400.000 LOGO programming 14.600.000 Haskell programming 250.000 [1] http://www.tiobe.com/index.php/content/paperinfo/tpci/index.html [2] http://en.wikipedia.org/wiki/Logo_(programming_language) [3] http://web.archive.org/web/20070606231519/www.tiobe.com/?tiobe_index -- Regards, Henk-Jan van Tuyl -- http://functor.bamikanarie.com http://Van.Tuyl.eu/ --

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Changing version numbering schemes for HackageDB packages?
by Corey O'Connor 13 Feb '09

13 Feb '09

I released a new version of data-spacepart that resolved some of the issues with the previous release. One issue I had was the previous release used the version numbering scheme I use at work: [date].[release] Which does not appear to work as well as the traditional X.Y.Z release numbering scheme with Cabal. As part of the new release I changed the version numbering scheme. An *obviously* bad idea if I thought it through. Any [date].[release] style version number is greater than a X.Y.Z version number until X gets rather large. So what to do? Continue using the [date].[release] version numbering scheme? Or is there a way to coax HackageDB to ignore the old release? EG: The new release is: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/data-spacepart-0… The old release is: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/data-spacepart-2… The "latest" release according to HackageDB is: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/data-spacepart Which points to the old release. Cheers, -Corey O'Connor

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How to deal with utf-8?
by Magicloud 13 Feb '09

13 Feb '09

Hi, I am using Text.CSV to read and using gtk2hs to display csv files using utf-8 encode. Well, it displays broken strings, seems like it cannot deal with utf-8. What should I do? Thanks.

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