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Postdoctoral Research Associate, "Refactoring Functional Programs"
by C.Reinke 07 Jan '02

07 Jan '02

Just a quick reminder that the closing date for this position, advertised here earlier, is *this Friday*, 11/01/2002. Happy New Year, Claus > Applications are invited for a three year Postdoctoral Research > Associate position at the University of Kent at Canterbury, to work > under the direction of Professor Simon Thompson and Dr Claus Reinke > on an EPSRC funded project entitled "Refactoring Functional > Programs". ... > Please telephone the Personnel Office for further particulars > on 01227 827837 (24 hours) or email: Personnel(a)ukc.ac.uk , quoting > the reference number R02/15. Text phone users please telephone > 01227 823674. The closing date is Friday 11 January 2002. > > Informal enquires can be directed to either Simon or Claus: > {S.J.Thompson,C.Reinke}@ukc.ac.uk. More details of the project > and further particulars for the position can be found at: > > http://www.cs.ukc.ac.uk/people/staff/sjt/Refactor/

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Update: Haskell IDE for Windows
by Rijk-Jan van Haaften 07 Jan '02

07 Jan '02

Hello, Recently, a new version of Hugs came out providing better graphics support. A long awaited feature. Several people asked me even before december to support the graphics library in JCreator as well, so an update of the JCreator add-on is now available. You can get it from http://www.students.cs.uu.nl/people/rjchaaft/JCreator/ (or via http://www.haskell.org/libraries/) There is one drawback: the Haskell Report is not available currently. I hope to fix it soon. For those who expect no news: I have significantly changed the readme.txt file, so please don't skip it. Regards, Rijk-Jan

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Re: The Wisdom of Time
by Ketil Z Malde 07 Jan '02

07 Jan '02

Keith Wansbrough <Keith.Wansbrough(a)cl.cam.ac.uk> writes: >> what is the wisdom behind representing a TimeDiff as a struct of year, >> month, week and so on, instead of simply the (fractional) number of >> seconds, or similar? > Firstly, I believe that the Time module is broken, and no one has yet > come up with a satisfactory design. Okay. > But the behaviour you describe is necessary. Consider: Right, thanks for the elaboration. What is the right way to calculate the time an operation takes? I thought just doing do t1 <- getClockTime putStrLn (show $ resultOfComputation) t2 <- getClockTime putStrLn (show $ diffClockTimes t2 t1) which seems to work well enough for my purposes. (I'd like to measure running times for various algorithms on various data sets, at the moment mainly to convince myself about time complexity, but perhaps later to compare against implementations on different compilers/languages. I could of course simply use /usr/bin/time, but it seems a bit of a hassle to compile each benchmark individually to do so.) -kzm -- If I haven't seen further, it is by standing in the footprints of giants

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The Wisdom of Time
by Ketil Z Malde 07 Jan '02

07 Jan '02

The following message is a courtesy copy of an article that has been posted to comp.lang.functional as well. Hi, what is the wisdom behind representing a TimeDiff as a struct of year, month, week and so on, instead of simply the (fractional) number of seconds, or similar? In particular, the derivation of Eq seems to me to imply that sixty minutes is not the same as one hour. I haven't gotten around to play much with the module yet, but at least Prelude> (Time.TimeDiff 0 0 0 1 0 0 0) == (Time.TimeDiff 0 0 0 0 60 0 0) False Could somebody please enlighten me on this issue? -kzm -- If I haven't seen further, it is by standing in the footprints of giants

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RE: Library Report Omission
by Simon Peyton-Jones 07 Jan '02

07 Jan '02

Reasonable suggstion, but I think it's too late to add new functionality. (Apart from anything else, we'd have to get into whether it could be implemented portable on all platforms. And H98 is really very stable now.) But we'll add it to the new IO libraries that are coming up for GHC/Hugs/nhc. Simon | -----Original Message----- | From: Ashley Yakeley [mailto:ashley@semantic.org] | Sent: 06 January 2002 06:02 | To: Haskell List | Subject: Library Report Omission | | | I'm not sure if I've mentioned this before... | | Module IO (sec. 11.4) is missing hSetFileSize. | | hSetFileSize :: Handle -> Integer -> IO () | | Or is this too big a change for the report at this stage? Without it | there's no way to shorten a file without deleting it and | rewriting it. | Very annoying... | | -- | Ashley Yakeley, Seattle WA | | | _______________________________________________ | Haskell mailing list | Haskell(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell |

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Library Report Omission
by Ashley Yakeley 06 Jan '02

06 Jan '02

I'm not sure if I've mentioned this before... Module IO (sec. 11.4) is missing hSetFileSize. hSetFileSize :: Handle -> Integer -> IO () Or is this too big a change for the report at this stage? Without it there's no way to shorten a file without deleting it and rewriting it. Very annoying... -- Ashley Yakeley, Seattle WA

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RE: Report Issues
by Simon Peyton-Jones 04 Jan '02

04 Jan '02

Yes, that's a good point. I think nub should be defined as nubBy (==), not as nubBy (not . (/=)). I'll make it consistently so. Simon | -----Original Message----- | From: Janis Voigtlaender [mailto:voigt@orchid.inf.tu-dresden.de] | Sent: 04 January 2002 11:07 | To: haskell(a)haskell.org | Subject: Report Issues | | | Simon Peyton-Jones wrote: | > | > Folks, | > | > You have all been eating too much Xmas pudding. Only one | > Haskell98 Report issue has arisen since my release of 21 Dec. | | OK, here comes a rather trivial issue regarding the libraries: | | Section 7.6 of the Library Report gives the following example | definition of nub: | | nub :: (Eq a) => [a] -> [a] | nub [] = [] | nub (x:xs) = x : nub (filter (\y -> x /= y) xs) | | But then, in Section 7.9, the following actual implementation | is given: | | nub :: Eq a => [a] -> [a] | nub = nubBy (==) | | nubBy :: (a -> a -> Bool) -> [a] -> [a] | nubBy eq [] = [] | nubBy eq (x:xs) = x : nubBy eq (filter (\y -> not | (eq x y)) xs) | | The two definitions are only equivalent, if for all x and y | holds that | (x /= y) | and | (not (x == y)) | are equivalent. | While this is true for all basic types, and is also true for | user defined instances of Eq, if the programmer specifies | only one of the two functions (==) or (/=) and leaves the | other one at the default method, nobody can prevent me from | writing an instance declaration where I define (==) and (/=) | without adhering to the duality. This might be stupid to do, | but still it contradicts the report, right? | | Janis. | | | -- | Janis Voigtlaender | http://wwwtcs.inf.tu-dresden.de/~voigt/ | mailto:voigt@tcs.inf.tu-dresden.de | | _______________________________________________ | Haskell mailing list | Haskell(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell |

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Storable Class
by Carolina Mattos Cavalcanti 04 Jan '02

04 Jan '02

Hi, I´m Carol, and I´m just need some help to find the predefined types from the storable class. I´ll be thankfull if you could answer.

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Please add me to the Haskell mailing list.
by Andrew Hamilton 04 Jan '02

04 Jan '02

Thanks. Regards, Andrew Hamilton tel: +44 (0) 7956 396396 email: ah93(a)zepler.org.uk

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Haskell 98 sections
by Simon Peyton-Jones 04 Jan '02

04 Jan '02

Folks, You have all been eating too much Xmas pudding. Only one Haskell98 Report issue has arisen since my release of 21 Dec. The issue is this: Section 3.5 says that (a + b +) is a valid operator section (meaning \x -> a + b + x) because (+) is right associative. But this is contradicted by the actual syntax, which only has the productions: aexp ::= '(' exp(i+1) qop(a,i) ')' | '(' qop(a,i) exp(i+1) ')' (Here 'a' ranges over the possible operator associativities of left, right, non-assoc, while i ranges over precendence levels.) This is clearly an inconsistency. I propose to fix the syntax in the way proposed by Simon and Ian (in a thread on GHC-bugs), by adding the following productions for aexp. '(' lexp(i) qop(l,i) ')' '(' qop(r,i) rexp(i) ')' This actually follows the way that qfunlhs is defined. There is no ambiguity, because an exp(i+1) is not an lexp(i). This should bring the formal syntax into line with the words in Section 3.5. Can anyone think of a reason why this won't work? Simon

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