Re[2]: [Haskell-cafe] deepSeq vs rnf
Hello Cale, Monday, October 23, 2006, 7:19:14 AM, you wrote: Speaking of boilerplate and the scrapping thereof, Data.Generics could theoretically also be used to write a relatively generic rnf/deepSeq, but in my attempts, it seems to be much much slower than using a specific normal form class. Here's my code from quite a while back. As I recall, it's semantically correct, but ran about an order of magnitude slower. There might be a much better way to do it, I don't really know Data.Generics all that well. by no means it's surprising - syb library works slower than hand-written or compile-time generated code because it implements _run-time_ polymorphism - data types are tested at run-time and then coerced to their actual types. there is also syb4 approach by Oleg where type classes used for polymorphism - it should have average speed, in theory -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re[2]: [Haskell] Pugs gains SMP parallelism support.
Hello Taral, Monday, October 23, 2006, 1:12:38 PM, you wrote: They probably are. However you get the overhead of creating the array (when you don't really need O(1) random access) and every thread signals the same semaphore which may lead to some congestion which could slow things down. You do need O(1) random access for the writers to put their results in efficiently. And newArray_ should be faster than N copies of newEmptyMVar. It is true that I have one congestion point (the semaphone) instead of N (the mvars). are you sure that writeArray calls may be overlapped? :) they may change some technical fields, such as 'dirty' bit -- Best regards, Bulatmailto:[EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: (more) type level functions (type of decreasing list)
[EMAIL PROTECTED] wrote: ] ] One way is to use existentials: ] data Seq1 a = forall b. (Pre a b, Show b) = Cons1 a (Seq1 b) | Nil1 ] ] which perhaps not entirely satisfactory because we have to specify all ] the needed classes in the definition of Seq1. Yes, that seems pretty burdensome, since we can't now create even a simple function like tail, right? Or at least I think existentials are the problem with this... tail1 :: (Pre a b, Show b) = Seq1 a - Seq1 b -- tail1 (Cons1 x xs) = xs tail1 Nil1 = error empty Seq ...ghci reports... Couldn't match expected type `b' (a rigid variable) against inferred type `b1' (a rigid variable) `b' is bound by the type signature for `tail_' at dec2.lhs:30:17 `b1' is bound by the pattern for `Cons1' at dec2.lhs:31:8-17 Expected type: Seq1 b Inferred type: Seq1 b1 In the expression: xs In the definition of `tail1': tail1 (Cons1 x xs) = xs ...and hugs... ERROR dec2.lhs:31 - Existentially quantified variable in inferred type *** Variable : _3 *** From pattern : Cons1 x xs *** Result type : Seq1 _0 - Seq1 _3 ] ] Perhaps a better -- and a more general alternative -- is to use ] type-level programming to its fullest. The following defines a list ] whose elements are constrained to be in the decreasing, increasing, ] or any other (defined in the future) order. This example shows that ] Haskell truly has more kinds than it is commonly acknowledged. snip ] data Cons a b = Cons a b ] data Nil = Nil ...if we build our lists with tuple-like Conses, our types end up being as long as our lists, right? So an infinite list has an infinite type, which seems like it could be problematic to type check. For example, here's an arbitrarly long increasing list... data Seq' a = Cons' a (Seq' (Succ a)) | Nil' deriving Show from :: a - Seq' a from x = Cons' x (from (S x)) inf = from Z ...with some of the usual functions... tail_ :: Seq' a - Seq' (Succ a) tail_ (Cons' x xs) = xs class Take a where take_ :: a - (Seq' b) - (Seq' b) instance Take Zero where take_ Z _ = Nil' instance Take a = Take (Succ a) where take_ (S n) (Cons' x xs) = Cons' x (take_ n xs) take_ _ Nil' = error can't take_ that many class Drop a b c | a b - c where drop_ :: a - Seq' b - Seq' c instance Drop Zero b b where drop_ Z xs = xs instance Drop a (Succ b) c = Drop (Succ a) b c where drop_ (S n) (Cons' x xs) = drop_ n xs drop_ _ Nil' = error can't drop_ that many ...Anyway, for fun, here's a list that alternates between two types in a linearly increasing manner. So you can start out with one Int and then two Strings, then three Ints, four Strings, etc. I don't know if it is going to yeild to a type classless GADT solution yet, but I'll keep thinking about it. data Fancy a b left next = forall c d left' next' lz decleft . (Show c, Show d, Zerop left lz, If lz (Succ next) next next', Pre left decleft, If lz next decleft left', If lz b a c, If lz a b d ) = Cons a (Fancy c d left' next') | Nil fancy :: Fancy Integer String Zero (Succ Zero) fancy = (Cons 1 (Cons a (Cons b (Cons 2 (Cons 3 (Cons 4 (Cons c (Cons d (Cons e (Cons f (Cons 5 (Cons 6 (Cons 7 (Cons 8 (Cons 9 Nil))) instance (Show a, Show b) = Show (Fancy a b c d) where show (Cons x xs) = (Cons ++ (show x) ++ ++ (show xs) ++ ) show Nil = Nil data Succ a = S a deriving Show data Zero = Z deriving Show data TTrue data TFalse class Pre a b | a - b instance Pre (Succ a) a instance Pre Zero Zero class If p t e result | p t e - result instance If TTrue a b a instance If TFalse a b b class Zerop a b | a - b instance Zerop Zero TTrue instance Zerop (Succ a) TFalse ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Exception: Too many open files
Hello Haskellers, I'm wondering how to get the following to work: I need to parse about 18.000 files. I would like to have a function 'parseFiles' that parses all these files and returns the result in a list. When I execute 'test' from the simplified code below I get the error: *** Exception: file_0014994.txt: openFile: resource exhausted (Too many open files) So it seems that 'parseFiles' tries to open all the ~18.000 files and gets exhausted when opening the 14994 file. What I would like is 'take 3 = parseFiles' to read only the first 3 files. Is this possible, and if so, what is the best way to do this? Note that the code below is a bit simplified: -- module Main where import Text.ParserCombinators.Parsec data T = ... test = print . take 3 = parseFiles parseFiles :: IO [T] parseFiles = mapM parseFile = getFileFPs getFileFPs :: IO [FilePath] getFileFPs = ... -- returns a large list of about 18.000 FilePaths parseFile :: FilePath - IO T parseFile fp = liftM getRight $ parseFromFile someParser fp getRight (Right r) = r someParser :: Parser T someParser = ... -- Greetings, Bas van Dijk ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] split string into n parts
I want to split a string into 5 parts of equal length, with the last fifth padded if necessary, but can't get it right - here's what I've got - fifths :: String - String fifths s = fifths' 0 s where l = (length s) `div` 5 fifths' xs c [] = xs ++ (replicate (l-c) 'X') fifths' xs c (y:ys) = if c == l then fifths' (xs++[' ',y]) 0 ys else fifths' (xs++[y]) (c+1) ys which, apart from surely being uglier than need be, doesn't work: *Main fifths IDOLIKETOBEBESIDETHESEASIDE IDOLI KETOBE BESIDE THESEA SIDEXX *Main fifths 12345 1 23 45 Any thoughts? Thanks! This isn't homework BTW, I'm having a go at the ruby quiz puzzles in haskell, which seems to be a nice way to learn. -- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6960346 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Exception: Too many open files
On Mon, Oct 23, 2006 at 08:48:24PM +0200, Bas van Dijk wrote: So it seems that 'parseFiles' tries to open all the ~18.000 files and gets exhausted when opening the 14994 file. What I would like is 'take 3 = parseFiles' to read only the first 3 files. Is this possible, and if so, what is the best way to do this? parseFiles :: IO [T] parseFiles = mapM parseFile = getFileFPs use this function instead of mapM above: unsafeInterleaveMapIO f (x:xs) = unsafeInterleaveIO $ do y - f x ys - unsafeInterleaveMapIO f xs return (y : ys) unsafeInterleaveMapIO _ [] = return [] Best regards Tomasz ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] split string into n parts
I want to split a string into 5 parts of equal length, with the last fifth padded if necessary, but can't get it right - here's what I've got - fifths :: String - String fifths s = fifths' 0 s where l = (length s) `div` 5 [... snip ...] Any thoughts? Thanks! This isn't homework BTW, I'm having a go at the ruby quiz puzzles in haskell, which seems to be a nice way to learn. Cool idea! Can you post a link for the puzzles? As for this one, don't you want the first multiple of five larger than the length of the string? You should be able to make things simpler if you auto-pad the string from the get-go (forgive any syntax errors, as I'm just composing in a browser here...): prepad :: Integer - String - String prepad n s | (length s) `mod` n == 0 = s prepad n s = prepad n (s ++ ' ') And now you can be sneaky: nths :: Integer - String - String nths n s = [first_nth n s] ++ nths (n-1) (after_first_nth n s) nths 1 s = s Where first takes the first n chars (e.g., take ((length s) `div` n)) and after_first_nth returns the tail of the list.. -- [EMAIL PROTECTED] http://mult.ifario.us/ -- [EMAIL PROTECTED] http://mult.ifario.us/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
jim burton [EMAIL PROTECTED] writes: I want to split a string into 5 parts of equal length, with the last fifth padded if necessary (snip) *Main fifths 12345 1 23 45 What's the correct answer for fifths 123456? I can't figure out how to meet both your constraints. Is 12 34 56 XX XX permitted (padding before fifth as well)? -- Mark ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
Paul Brown-4 wrote: Cool idea! Can you post a link for the puzzles? Thankyou! It's http://www.rubyquiz.com - They are mostly well suited to haskell, lot of mazes etc. I've done 5 or 6 with varying degrees of success but have learned a lot. This thing about strings in fifths is from #1, the solitaire cipher. As for this one, don't you want the first multiple of five larger than the length of the string? You should be able to make things simpler if you auto-pad the string from the get-go (forgive any syntax errors, as I'm just composing in a browser here...): [snip] Good plan! I will play with that. -- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6961186 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
jim burton [EMAIL PROTECTED] writes: (snip) *Main fifths IDOLIKETOBEBESIDETHESEASIDE IDOLI KETOBE BESIDE THESEA SIDEXX *Main fifths 12345 1 23 45 (snip) FWIW this unholy thing works for me, fifths :: String - String fifths = splitIntoN 5 splitIntoN :: Int - String - String splitIntoN n string = let stringToSplit = string ++ replicate (n-1) 'X' in unwords (map fst (take n (tail (iterate (splitAt (div (length stringToSplit) n) . snd) (undefined, stringToSplit) Admittedly, a 'let' might be nice to name some intermediate computations. -- Mark ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
Mark T.B. Carroll-2 wrote: FWIW this unholy thing works for me, fifths :: String - String fifths = splitIntoN 5 [snip] Thanks Mark. -- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6961461 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
On 10/23/06, jim burton [EMAIL PROTECTED] wrote: I want to split a string into 5 parts of equal length, with the last fifth padded if necessary, but can't get it right I got this:fifths :: String - Stringfifths xs = let len = (length xs + 4) `div` 5 padded = take (len * 5) (xs ++ ) in unwords $ nth len padded where nth _ [] = [] nth n xs = (take n xs) : (nth n $ drop n xs) *Main fifths IDOLIKETOBEBESIDETHESEASIDE IDOLI KETOBE BESIDE THESEA SIDEXX *Main fifths 12345 1 23 45This gives the following results:IDOLIK ETOBEB ESIDET HESEAS IDE and1 2 3 4 5But it also gives this result, which may or may not be correct for your problem: *Main fifths 12345612 34 56 -- RichAIM : rnezzyICQ : 174908475 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: split string into n parts
jim burton [EMAIL PROTECTED] writes: Paul Brown-4 wrote: Cool idea! Can you post a link for the puzzles? Thankyou! It's http://www.rubyquiz.com - They are mostly well suited to haskell, lot of mazes etc. I've done 5 or 6 with varying degrees of success but have learned a lot. This thing about strings in fifths is from #1, the solitaire cipher. At a quick glance I can't see which bit needs it. The only mention of five is where it asks to split the string into groups of five characters (not into five equal parts), padded with Xs. You can do that like this: splitAtMb n l = let p = splitAt n l in if null $ fst p then Nothing else Just p in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (length l `mod` 5) 'X') To break a string into five equal parts with the last padded with Xs, try this: fifths l = let len = length l part_length = (len+4)`div`5 pad_length = 5*part_length - len in unfoldr (splitAtMb part_length) (l ++ replicate pad_length 'X') I haven't checked these at all carefully, but at least they illustrate the use of unfoldr. [aside: One might argue that the prelude ought to provide splitAtMb rather than splitAt.] -- Jón Fairbairn [EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
Jón Fairbairn-2 wrote: At a quick glance I can't see which bit needs it. The only mention of five is where it asks to split the string into groups of five characters (not into five equal parts), padded with Xs. Oh dear, you're right. Sorry, I read in a rush. Thanks for the solution too. Jón Fairbairn-2 wrote: You can do that like this: splitAtMb n l = let p = splitAt n l in if null $ fst p then Nothing else Just p in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (length l `mod` 5) 'X') To break a string into five equal parts with the last padded with Xs, try this: fifths l = let len = length l part_length = (len+4)`div`5 pad_length = 5*part_length - len in unfoldr (splitAtMb part_length) (l ++ replicate pad_length 'X') I haven't checked these at all carefully, but at least they illustrate the use of unfoldr. [aside: One might argue that the prelude ought to provide splitAtMb rather than splitAt.] -- Jón Fairbairn [EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe -- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6961825 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
tweak to in_fives in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (5 - length l `mod` 5) 'X') -- View this message in context: http://www.nabble.com/split-string-into-n-parts-tf2496941.html#a6961912 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: split string into n parts
jim burton [EMAIL PROTECTED] writes: tweak to in_fives in_fives l = unfoldr (splitAtMb 5) (l ++ replicate (5 - length l `mod` 5) 'X') Whoops! Yes. And a slapped wrist for me for writing a constant three times. Serves me right for not writing groups_of n l = unfolder (splitAtMb n) ... in_fives = groups_of 5 :-) -- Jón Fairbairn [EMAIL PROTECTED] ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Exception: Too many open files
On Monday 23 October 2006 21:50, Tomasz Zielonka wrote:
unsafeInterleaveMapIO f (x:xs) = unsafeInterleaveIO $ do
y - f x
ys - unsafeInterleaveMapIO f xs
return (y : ys)
unsafeInterleaveMapIO _ [] = return []
Great it works! I didn't know about unsafeInterleaveIO [1].
Why is it called 'unsafe'?
And how does the laziness work? Is it because of the 'let r = ... in (s, r)'?
[1] ghc/libraries/base/GHC/IOBase.lhs:249:
---
{-|
'unsafeInterleaveIO' allows 'IO' computation to be deferred lazily.
When passed a value of type @IO a@, the 'IO' will only be performed
when the value of the @a@ is demanded. This is used to implement lazy
file reading, see 'System.IO.hGetContents'.
-}
{-# INLINE unsafeInterleaveIO #-}
unsafeInterleaveIO :: IO a - IO a
unsafeInterleaveIO (IO m)
= IO ( \ s - let
r = case m s of (# _, res #) - res
in
(# s, r #))
-- We believe that INLINE on unsafeInterleaveIO is safe, because the
-- state from this IO thread is passed explicitly to the interleaved
-- IO, so it cannot be floated out and shared.
---
Thanks,
Bas van Dijk
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[Haskell-cafe] Guards with do notation?
hello all, why is it not possible to use guards in do-expressions like do (a, b) | a == b - getPair return a and b are equal Cheers, Misha ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Segmentation fault with realdine library
Hi, This small program says Segmentation fault: module Main (Main.main) where import Data.Char import System.Time import System.Console.Readline main :: IO () main = do readKey return () I don't understand anything about readline, so I probably should have to call some function before readKey (by they way, I don't know what readKey does). But I thought a segmentation fault is always something to be reported. Best, Maurício ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
jim burton wrote: I want to split a string into 5 parts of equal length, with the last fifth padded if necessary, but can't get it right - here's what I've got - fifths s = unwords.take 5.unfoldr (Just . splitAt l) $ s ++ repeat ' ' where l = (length s + 4) `div` 5 Of course no Haskeller in his right mind would carelessly apply the final 'unwords' unless this was for immediate output. Udo. -- The Second Law of Thermodynamics: If you think things are in a mess now, just wait! -- Jim Warner signature.asc Description: Digital signature ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Segmentation fault with realdine library
On Mon, 2006-10-23 at 20:18 -0200, Maurício wrote: Hi, This small program says Segmentation fault: module Main (Main.main) where import Data.Char import System.Time import System.Console.Readline main :: IO () main = do readKey return () I don't understand anything about readline, so I probably should have to call some function before readKey (by they way, I don't know what readKey does). But I thought a segmentation fault is always something to be reported. Yes, I get the same result. It seems that one must use 'initialise' first. I agree, it would be much nicer if you didn't have to do this and if instead it would Just Worktm. Duncan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Exception: Too many open files
On Tuesday 24 October 2006 00:58, Greg Fitzgerald wrote: test = print . take 3 = parseFiles I haven't had time to double-check this code, but something like this ought to work (no 'unsafe' operations!): test = sequence . take 3 . map (print . parseFile) = getFileFPs Let me know how it goes. Indeed, the following works: test = sequence . take 3 . map (\fp - print = parseFile fp) = getFileFPs Thanks, Bas. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Read a single char
Hi, How can I read a single character from standard output? I would like the user to press a single key and the reading function return imediately after that key is pressed. Thanks, Maurício ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] split string into n parts
Udo Stenzel wrote: jim burton wrote: I want to split a string into 5 parts of equal length, with the last fifth padded if necessary, but can't get it right - here's what I've got - fifths s = unwords.take 5.unfoldr (Just . splitAt l) $ s ++ repeat ' ' where l = (length s + 4) `div` 5 Okay, you win. That's the nicest answer so far, I think. But here are solutions with a different theme altogether. They are based on groupBy, not unfoldr. I really like the new `on` function. module Chunk where import Data.List (on) f g = \x y - f (g x) (g y) groupByIndex test xs = map (map snd) $ groupBy (test `on` fst) $ zip [0..] xs -- chunk : divide the input string into n chunks of equal length (len), with padding -- chunk1 accepts the number of chunks chunk1 n pad xs = unwords $ take n $ groupByIndex ((==) `on` (`div` len)) $ xs ++ repeat pad where len = (length xs + n - 1) `div` n -- chunk2 accepts the length of each chunk chunk2 len pad xs = unwords $ take n $ groupByIndex ((==) `on` (`div` len)) $ xs ++ repeat pad where n = (length xs + len - 1) `div` len ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Read a single char
briqueabraque: Hi, How can I read a single character from standard output? I would like the user to press a single key and the reading function return imediately after that key is pressed. so you want a function of type: IO Char asking Hoogle (http://haskell.org/hoogle) we get: Prelude. getChar:: IO Char IO. hGetChar :: Handle - IO Char -- Don ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Read a single char
Hi getChar doesn't return until I press Enter. I need something that returns immediately after I press any key. It's a problem with buffering: http://haskell.org/hoogle/?q=buffering suggests: hSetBuffering stdin NoBuffering Thanks Neil ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Re: Read a single char
On 10/23/06, Neil Mitchell [EMAIL PROTECTED] wrote: HigetChar doesn't return until I press Enter. I need something that returns immediately after I press any key.It's a problem with buffering:hSetBuffering stdin NoBuffering This usually doesn't work on Windows:GHC 6.4.2 and 6.6: requires enterHugs (console) Sept. 2006: requires enter WinHugs (GUI) Sept. 2006: works as expectedBut it seems to work on Linux:GHC 6.4.1 on Ubuntu 6.06: works as expectedGHC 6.6 on Ubuntu 6.06: works as expectedI am really interested in hearing of a solution that works on all platforms. import IO main = do hSetBuffering stdin NoBuffering hGetChar stdinRegards,Brian ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
