January 2009 - Beginners

Function Type Confusion ..
by Tom Poliquin 28 Jan '09

28 Jan '09

I was reading "Arrows and Computation" http://www.soi.city.ac.uk/~ross/papers/fop.ps.gz (trying to lose my 'beginner' status) when I saw (on page one) add :: (b -> Int) -> (b -> Int) -> (b -> Int) add f g b = f b + g b It seemed like the type definition was wrong (short at least). I tried it anyway .. module Main where add :: (b -> Int) -> (b -> Int) -> (b -> Int) add f g b = f b + g b main = do x <- return $ add (+2) (+3) 7 print x The program compiles and runs and produces '19' ! For fun I loaded into ghci and got what I believe is the proper type .. *Main> :t add add :: (b -> Int) -> (b -> Int) -> b -> Int When I try the same thing with something simpler (leaving a bit off the type definition) I get the expected error (by me) .. module Main where dog :: Int -> Int dog a b = a + b main = do x <- return $ dog 2 3 print x Main.hs:5:0: The equation(s) for `dog' have two arguments, but its type `Int -> Int' has only one What am I missing? .. Apparently something fundamental about type definitions .. Any help appreciated. Tom

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Follow up to reference request
by Alan Cameron 27 Jan '09

27 Jan '09

Hi, I am indebted to Andrew Wagner for pointing me in the direction of the book http://book.realworldhaskell.org/read/ Which I now have time to study. Everything was going fine until I got to the section in Getting Started where it gives A Simple Program. Now as someone who is used to the Windows environment CLI is not my forte. Following instructions which appear to be incomplete I created a file of the program WC.hs but where should it go? I tried various places until I reread the :? For the :cd command. Ah I said that's what I need to do put it in a folder and change the Dir. Now I can find the file with the :edit command. Did the same with the file quux.txt same place. Now run the command $ runghc WC < quux.txt <interactive>:1:0: parse error on input '$' What have I done wrong or not done? It does say "at a shell or command prompt" what's that?? I reverted to the installation instructions provided by GHC for Windows. 2.2 Installing on Windows 2.2.1 Installing GHC on Windows Checked all that had been done correctly. Enter the program line 1> bash$ cat main.hs <interactive>:1:0: parse error on input '$' Now bash looks like a modified prompt?? So what is happening? Regards to all and TIA for reply. Alan Cameron

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Calculating Time Complexity
by Matthew J. Williams 26 Jan '09

26 Jan '09

Dear all, In general terms, how would one calculate the time complexity of a given algorithm? Feel free to make use of my pseudo code in your answer: /* input: 2-D array A of size n by n output: a number max */ Max := 0 For i := 1 to n sum := 0 For j := 1 to n sum := sum + A[i][j] End for If sum > max then max := sum End for Output max Sincerely Matthew J. Williams

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deleteM :: (Ord a) => a -> Maybe (Set a) -> Maybe (Set a)
by Martin Hofmann 26 Jan '09

26 Jan '09

I often come across the problem to insert into a collection which might not exist yet, or delete from it and want the collection to be deleted if it is empty afterwards. I always end up with these two functions: deleteM :: (Ord a) => a -> Maybe (Set a) -> Maybe (Set a) deleteM e s = liftM (S.delete e) s >>= \s' -> if S.null s' then return s' else Nothing insertM :: (Ord a) => a -> Maybe (Set a) -> Maybe (Set a) insertM e s = case s of (Just s') -> return $ S.insert e s' Nothing -> return $ S.insert S.empty Is there a way to express each in a (polymorphic) point-free one-liner? If not, why isn't there such a function for the standard collection, because IMHO this is what you need when using 'alter'. Thanks, Martin

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($) operator
by John Hartnup 26 Jan '09

26 Jan '09

Hi. I'm working through Real World Haskell, and although it's going well (I just finished the exercise to write a glob matcher without using a regex library, and I'm pleased as punch), I keep seeing the ($) operator, and I'm not sure I understand its use. If the book explains it, I've been unable to find it. Empirically, it seems like: a $ b c d e f .. is equivalent to .. a (b c d e f) But is that it's only purpose? To placate the LISP haters by removing parentheses? (1 +) 2 does the same thing as (1 +) $ 2, and has the same type. Am I missing something? Thanks, John -- "There is no way to peace; peace is the way"

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Re: [Haskell-beginners] ($) operator
by Brent Yorgey 26 Jan '09

26 Jan '09

On Sun, Jan 25, 2009 at 05:34:35AM +0000, John Hartnup wrote: > 2009/1/24 Brent Yorgey <byorgey(a)seas.upenn.edu>: > > > > > map ($5) [(+1), (^2), abs] > > > > which evaluates to [6,25,5]. > > > > Just to check I understand correctly - you're using ($) to 'promote' > an expression into a function, in order that it may be used in higher > order functions? In this case, I'm using ($) to *make* a higher-order function---($5) is a function which takes its argument (which must be a function) and applies it to 5. (I'm assuming you're familiar with 'operator sections', if you aren't I can explain that too.) -Brent

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[fschwidom@localhost: Re: [Haskell-beginners] ($) operator]
by Frank Schwidom 25 Jan '09

25 Jan '09

----- Forwarded message from Frank Schwidom <fschwidom@localhost> ----- Date: Sat, 24 Jan 2009 21:40:54 +0000 From: Frank Schwidom <fschwidom@localhost> To: John Hartnup <john.hartnup(a)gmail.com> Subject: Re: [Haskell-beginners] ($) operator Message-ID: <20090124214053.GA4496@BigBox> References: <6c1ba2fc0901241137x242cf4b4w2398896af45ae694(a)mail.gmail.com> MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: <6c1ba2fc0901241137x242cf4b4w2398896af45ae694(a)mail.gmail.com> User-Agent: Mutt/1.5.16 (2007-06-09) Status: RO On Sat, Jan 24, 2009 at 07:37:29PM +0000, John Hartnup wrote: > Hi. > > I'm working through Real World Haskell, and although it's going well > (I just finished the exercise to write a glob matcher without using a > regex library, and I'm pleased as punch), I keep seeing the ($) > operator, and I'm not sure I understand its use. If the book explains > it, I've been unable to find it. > > Empirically, it seems like: > a $ b c d e f > .. is equivalent to .. > a (b c d e f) > > But is that it's only purpose? To placate the LISP haters by removing > parentheses? > > (1 +) 2 does the same thing as (1 +) $ 2, and has the same type. > > Am I missing something? > > Thanks, > John > > > -- > "There is no way to peace; peace is the way" > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners maybe can this answer your question: A: ((==) True) (1 == 1) -- won't work: ((==) True) 1 == 1 => ((==) True) ((==) 1 1) -- won't work: ((==) True) (==) 1 1 => ((==) True) $ (==) 1 1 => ((==) True) $ 1 == 1 Regards ----- End forwarded message -----

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Type classes are not like interfaces, after all
by Francesco Bochicchio 24 Jan '09

24 Jan '09

Hi all. It's about a month I'm trying to learn haskell in my spare time ( and, I should add, with my spare neuron :-). I made progress, but more slowly than I expected :-(. I started hasking question on comp.lang.haskell (I like newsgroups more than mailing lists), but then realized that this may be a better places for my newbie questions. So, here comes the first ... As many beginners coming from OOP, i made the mental equation 'type class == interface'. It worked well ... unitil yesterday. Then I discovered that this piece of code (1) is illegal in askell (ghc gives the 'rigid type variable' error) Num n => a :: n a = 3 :: Integer I also discovered that this (2) instead is legal: Num n => a :: n a = 3 because it is implicitely translated into (3): Num n => a :: n a = fromInteger 3 But I don't understand the difference between (1) and (3). I could not find the implementation of fromInteger, but I suspect that it goes like this: instance Num Integer where fromInteger x -> x To me, it looks like (1) and (3) are equal, but that the compiler needs a sort of 'formal reassurance' that the value yielded by 'a' is of Num type. But since Integer is an instance of Num, I expected that (1) was already giving this reassurance. At least, it works this way with interfaces in Java and with class pointers in C++. But obviously there is something I don't get here ... Anybody here can explain? Ciao ----- FB

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case in monadic function
by Kovacs David 24 Jan '09

24 Jan '09

Hello! I have a function like this: http://rafb.net/p/8E66FI29.html, starting with evalState. The problem is with checkPredName cause it's return type is SemanticError what's not a monadic value but case waiting for m a (as the error message sais), but if I use return to checkPredName then ofc the pattern match will fail. How can I fix this? Regards, David

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Final call for papers: MULTICONF-09
by John Edward 24 Jan '09

24 Jan '09

The 2009 Multi Conference in Computer Science, Information Technology and Control systems and Computational Science and Computer Engineering (MULTICONF-09) (website: http://www.PromoteResearch.org) will be held during July 13-16 2009 in Orlando, FL, USA. We invite draft paper submissions. The event consists of the following conferences: · International Conference on Artificial Intelligence and Pattern Recognition (AIPR-09) · International Conference on Automation, Robotics and Control Systems (ARCS-09) · International Conference on Bioinformatics, Computational Biology, Genomics and Chemoinformatics (BCBGC-09) · International Conference on Enterprise Information Systems and Web Technologies (EISWT-09) · International Conference on High Performance Computing, Networking and Communication Systems (HPCNCS-09) · International Conference on Information Security and Privacy (ISP-09) · International Conference on Recent Advances in Information Technology and Applications (RAITA-09) · International Conference on Software Engineering Theory and Practice (SETP-09) · International Conference on Theory and Applications of Computational Science (TACS-09) · International Conference on Theoretical and Mathematical Foundations of Computer Science (TMFCS-09) The website http://www.PromoteResearch.org contains more details. Sincerely John Edward Publicity committee

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