January 2006 - Haskell-Cafe

How to print a string (lazily)
by Daniel Carrera 04 Jan '06

04 Jan '06

Hello, I've been studying more Haskell and I've improved a lot. But I just hit a small problem. I want to print all the elements of a linst (putStr). I'd like to write something like this: print_list [] = do putStr "" print_list (x:xs) = (do putStr x) && print_list xs I know this is wrong, but I hope you can see what I'm trying to do. I know of other ways I could print a list. For example: print_list xs = do putStr(join xs) where join [] = "" join (x:xs) = (show x) ++ "\n" ++ join xs But the thing is, I want to write a lazy version of this function. It's not that I need such a function, I'm just trying to learn Haskell. Any suggestions? Question: What do you call a function that has side-effects? (like putStr) I know that "function" is the wrong term. Cheers, Daniel. -- /\/`) http://oooauthors.org /\/_/ http://opendocumentfellowship.org /\/_/ \/_/ I am not over-weight, I am under-tall. /

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Re: Progress on shootout entries
by Kimberley Burchett 04 Jan '06

04 Jan '06

I took a quick crack at optimizing fannkuch.hs. I got it down from 33s to 1.25s on my machine, with N=9. That should put it between forth and ocaml(bytecode) in the shootout page. The main changes I made were using Int instead of Int8, foldl' to accumulate the max number of folds, a custom flop function rather than a combination of reverse and splitAt, and a simpler definition for permutations. http://kimbly.com/code/fannkuch.hs Kimberley Burchett

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RE: Haskell vs. Erlang: The scheduler
by Simon Marlow 03 Jan '06

03 Jan '06

On 03 January 2006 15:13, Joel Reymont wrote: > On Jan 3, 2006, at 2:30 PM, Simon Marlow wrote: >> The default context switch interval in GHC is 0.02 seconds, >> measured in CPU time by default. GHC's scheduler is stricly round- >> robin, so therefore with 100 threads in the system it can be 2 >> seconds between a thread being descheduled and scheduled again. >> >> I measured the time taken to unpickle those large 50k packets as >> 0.3 seconds on my amd64 box (program compiled *without* >> optimisation), so the thread can get descheduled twice during while >> unpickling a large packet, giving a >4s delay with 100 threads >> running. > > Is it impractical then to implement this type of app in Haskell? > Based on the nature of Haskell scheduling I would be inclined to say > yes. Absolutely not! Apart from the problem you have with a space leak caused by the input buffer of the logging thread overflowing, which is easily fixed by using a bounded channel, I don't know why you want priorities. Is there something else? We could easily add Erlang-style priorities (though without the overhead of counting function calls, I'd do it by counting allocations), but I'd rather not if we can avoid it. Cheers, Simon

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file i/o
by Robert Heffernan 03 Jan '06

03 Jan '06

Hi, I am relatively new to Haskell and am finding i/o difficult to work with. I am trying to do something like the following: I have a file of data, each line of which looks like this: <STRING>, INTEGER SEQUENCE for example: FOO ,2,1,4,3,6,7,5,9,10,11,8,13,12, I would like to write a function the reads this file and returns arrays like this: ["FOO",[2,1,4,3,6,7,5,9,10,11,8,13,12]] Ideally, the function would return the first line when initially called, the second the next time it is called and so on. I would settle for something that returned a big array comprising arrays of the above type containing all the information in the file. The file is big, however. I can not figure out how to do this from any of the tutorials so I thought I might ask here. Thank you for your help, Robert.

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Joels Time Leak
by Adrian Hey 03 Jan '06

03 Jan '06

Hello, I haven't followed everything that's happened on the Binary IO thread, but has anybody else actually tried Joels code? .. http://wagerlabs.com/timeleak.tgz I can reproduce the problem (ghc/Linux), but can't explain it. It seems very strange that friggin about with an otherwise irrelevant (AFAICT) MVar fixes the problem. Regards -- Adrian Hey

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Re: Joels Time Leak
by Chris Kuklewicz 03 Jan '06

03 Jan '06

Thanks for the answer, but I should I written a longer comment. I have added such a longer comment below: Simon Marlow wrote: > Chris Kuklewicz wrote: > >> Another comment: between 1000's of threads and writing a custom >> continuation based scheduler, what about using a thread pool? Does >> anyone have a library with a "fork-IO-Pool" command? > > > You don't need a thread pool, because threads are so cheap. Thread > pools are just a workaround for lack of lightweight concurrency. > > Cheers, > Simon > Since the round-robin scheduler has (0.02 * N) seconds of delay for N therads, then one could trade off latency between time spent waiting for the thread pool to start a job and time spend running the job and getting interrupted. In the limit of 1 worker thread, all the latency is waiting to get run, and there are no interruptions, so the time taken *while running* is very short. With 10 threads, there can be a delay to start, and each interruption adds 0.2 seconds to the job's run time once it has started. For a server, the client requests queue up and wait for room in the thread pool, and the pool is kept small enough that the round-robin-schedular-delay keeps requests from timing out while being serviced. Otherwise 1000 client requests would cause 20 seconds of reschedule penalty for all threads and they could all timeout. With a thread pool, one can drop threads that have been waiting for too long instead of running them, so those threads will timeout. But the pool keeps servicing at least some of the client requests on time. All hypothetical to me, of course. -- Chris

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Haskell vs. Erlang: Logging to one thread from thousands /Answer/
by Joel Reymont 03 Jan '06

03 Jan '06

I asked the Erlang guys why I can log to a single process in Erlang without any problems. The scheduler could well be round-robin but since the message queue is hard-wired to each Erlang process they found an elegant way out. -- There is a small fix in the scheduler for the standard producer/consumer problem: A process that sends to a receiver having a large receive queue gets punished with a large reduction (number of function calls) count for the send operation, and will therefore get smaller scheduling slots. -- Thanks, Joel -- http://wagerlabs.com/

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Help with "shootout"
by Chris Kuklewicz 03 Jan '06

03 Jan '06

Hello all, I decided to put together an entry for one of the shooutout categories: http://haskell.org/hawiki/ChameneosEntry It involves 4 threads. There is no current Haskell entry. Anyone have comments? Alterations? I can't have created the best code yet. I don't have the compilers for the other entries, so I can't compare speed on my powerbook (which is the wrong kind of CPU anyway -- they test on an AMD system). -- Chris

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RE: [Haskell-cafe] ST monad
by Ralf Lammel 03 Jan '06

03 Jan '06

... for the same reason as this one doesn't get through: import Control.Monad.ST import Data.Array.ST main = print $ runST $ do return () ... but this one does: import Control.Monad.ST import Data.Array.ST main = print $ runST ( do return ()) it's all about rank-2 types; see SPJ's et al. paper on type inference for these types. However, I guess that the jury is still out, say this specific rank-2 behavior may be revised (and I also hope so). HTH Ralf > -----Original Message----- > From: haskell-cafe-bounces(a)haskell.org [mailto:haskell-cafe- > bounces(a)haskell.org] On Behalf Of Bulat Ziganshin > Sent: Tuesday, January 03, 2006 2:28 AM > To: haskell-cafe(a)haskell.org > Subject: [Haskell-cafe] ST monad > > Hello > > the following code can't go through typechecking. can anyone help me > to fix it or, better, let me know what i need to read to fix it myself? :) > > import Control.Monad.ST > import Data.Array.ST > main = print $ runST $ > do arr <- newArray (1,10) 127 > a <- readArray arr 1 > writeArray arr 1 216 > b <- readArray arr 1 > return (a,b) > > > PS: error message is > > b.hs:4:15: > Inferred type is less polymorphic than expected > Quantified type variable `s' escapes > Expected type: ST s a -> b > Inferred type: (forall s1. ST s1 a) -> a > In the first argument of `($)', namely `runST' > In the second argument of `($)', namely > `runST > $ (do > arr <- newArray (1, 10) 127 > a <- readArray arr 1 > writeArray arr 1 216 > b <- readArray arr 1 > return (a, b))' > > > > -- > 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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ST monad
by oleg＠pobox.com 03 Jan '06

03 Jan '06

Bulat Ziganshin wrote: the following code can't go through typechecking > import Control.Monad.ST > import Data.Array.ST > main = print $ runST $ > do arr <- newArray (1,10) 127 > a <- readArray arr 1 > writeArray arr 1 216 > b <- readArray arr 1 > return (a,b) Indeed. The short answer: use runST (long expression) rather than runST $ long expression when it comes to higher-ranked functions such as runST. A longer answer: http://www.haskell.org/pipermail/haskell-cafe/2004-December/008062.html > let me know what i need to read to fix it myself MLF (see Daan Leijen, A. Loeh, `Qualified types for MLF', ICFP05)

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