March 2010 - Beginners

Better way to write this?
by Tim Perry 09 Mar '10

09 Mar '10

I'm working my way through the 99 sample programs for haskell and I'm on #4: find the number of elements in a list. http://haskell.org/haskellwiki/99_questions/1_to_10 I wrote the obvious recursion. Then I rewrote it using foldl myLength :: [a] -> Int myLength [] = 0 myLength xs = foldl addOne 0 xs where addOne lhs rhs = lhs + 1 However, the solution given uses notation that confuses my little mind. Can someone point me to a good resource on this notation? myLength :: [a] -> Int myLength = foldr (\x n -> n + 1) 0 Thanks, Tim

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Re: Printing the result of a function evaluation
by Travis Erdman 09 Mar '10

09 Mar '10

Steve, I should have provided the following details ... The code I pasted was a timing function I've come across in various Haskell literature, and it "works" as is ... at least, the timing results it prints are plausible. My sole "contribution" to the code was the "print v" line, in an attempt to have it print BOTH the result of the function evaluation AND the time it took to evaluate it (and still only evaluating the function once). As noted, my "print v" prints only "()" regardless of the function -- not what I wanted. But the timing part still works. There has got to be an easy fix here, but I have yet to discover it ...

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Re: [Haskell-beginners] Help in pattern matching
by Stephen Tetley 08 Mar '10

08 Mar '10

Hi Joe - I've cc'ed back to list... I'm supposing from what you've said that only DataConst1 & DataConst2 have ids - so I've made a type synonym VId type VId = Int data TestData = DataConst1 VId String | DataConst2 VId Int | DataConst3 Sting String | DataConst4 String Int If both DataConst1 & DataConst2 have VId's the you right - you have to pattern match for both: setVid_to_4 :: TestData -> TestData setVid_to_4 (DataConst1 _ y) = DataConst1 4 y setVid_to_4 (DataConst2 _ y) = DataConst2 4 y setVid_to_4 var = var If all the cases of TestData had a VId e.g.: data TestData = DataConst1 VId String | DataConst2 VId Int | DataConst3 VId Sting String | DataConst4 Vid String Int ... then you could factor out VId to get these two data types data FactoredTD = FactoredTD VId TD data TD = DC1 String | DC2 Int | DC3 String String | DC4 String Int setVid_to_4 :: FactoredTD -> FactoredTD setVid_to_4 (FactoredTD _ y) = FactoredTD 4 y This factoring transformation is quite common. For your particular data type you could factor the other way: data TestDataX = DataConst1_or_2 VId (Either String Int) | DataConst3 Sting String | DataConst4 String Int setVid_to_4 :: TestDataX -> TestDataX setVid_to_4 (DataConst1_or_2 _ y) = DataConst1_or_2 4 y setVid_to_4 y = y However this factoring is quite horrible - I can't think of anywhere where I've seen it. Generally I wouldn't be too concerned about redundancy in pattern matching. If its easy to consistently name all the constructors in an algebraic data type, then that's a strong indication that you've put it into a good form. If you're still taxed by duplicate pattern matching you can then write projection and modification functions to do it once and once only: -- Has to return a Maybe as DC3 and DC4 have no ID getVId :: TestData -> Maybe VId getVid (DataConst1 iden _) = Just iden getVid (DataConst2 iden _) = Just iden getVid _ = Nothing -- note - uses an update function that accesses the original value -- rather than a simple replace -- updateVid :: (Vid -> Vid) -> TestData -> TestData updateVid f (DataConst1 iden y) = DataConst1 (f iden) y updateVid f (DataConst2 iden y) = DataConst2 (f iden) y updateVid f var = var incrementVid :: TestData -> TestData incrementVid = updateVid (\x -> x+1) Best wishes Stephen On 8 March 2010 12:19, Joe Fox <fox.joe87(a)gmail.com> wrote: > Hi Guys, > > First thanks for all yours inputs. > > Actually test is just a simple function , actually my actual function is > pretty complex and huge ... > > let me redefine Test function , i guess i took a bad example which created > lot of confusion > > the type of the test functions will be something like this , (i don't have > any ambiguity on type of the function, ) > > test:: TestData -> TestData > > My question is like is there any way i can match multiple pattern in a > function statement . > > like this (I know this doesnt work , but just trying to explain whats on my > mind) > > > test var@(DataConst1 x y) == var@(DataConst2 x y) = var{id=4} -- let's say > id is the record holder for first record > test var@(_) = var --- any for other two its returns the same > > the working code would be like > test var@(DataConst1 x y) = var{id=4} > test var@(DataConst2 x y) = var{id=4} --- I want these two lines in one > line, because the logic is same > test var@(_) = var > > some thing of this sort. > I hope i didnt add to the confusion > > > Thanks > Joe >

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Printing the result of a function evaluation
by Travis Erdman 08 Mar '10

08 Mar '10

Why oh why doesn't this work as I expect, and what is the simplest fix? speedTest 4 = 44 time :: (Show t) => IO t -> IO t time a = do start <- getCPUTime v <- a end <- getCPUTime let diff = (fromIntegral (end - start)) / (10^12) print v -- WHY DOESNT THIS WORK? ALWAYS PRINTS () .... printf "Computation time: %0.3f sec\n" (diff :: Double) return v main :: IO () main = do args <- getArgs time $ speedTest (read $ head args) `seq` return ()

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Call stack
by Nicolas Couture-Grenier 08 Mar '10

08 Mar '10

Is there an easy way to print the call stack of the evaluation of an expression in ghci without setting breakpoints or changing the code? That would be very useful.

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Speed performance problem on Windows?
by Travis Erdman 08 Mar '10

08 Mar '10

I'm working through one of Don Stewart's many excellent articles ... http://cgi.cse.unsw.edu.au/~dons/blog/2008/06/04#fast-fusion I faithfully re-created the source of his initial GHC reference implementation as: import System.Environment import Text.Printf mean :: Double -> Double -> Double mean n m = go 0 0 n where go :: Double -> Int -> Double -> Double go s l x | x>m = s / fromIntegral l | otherwise = go (s+x) (l+1) (x+1) main = do [d] <- map read `fmap` getArgs printf "%f\n" (mean 1 d) Then, compiled and executed like this: C:\Documents and Settings\Travis\My Documents\Haskell Code>ghc -O2 biglistmean.hs -optc-O2 -fvia-C --make -fforce-recomp [1 of 1] Compiling Main ( biglistmean.hs, biglistmean.o ) Linking biglistmean.exe ... C:\Documents and Settings\Travis\My Documents\Haskell Code>biglistmean 1000000 500000.5 C:\Documents and Settings\Travis\My Documents\Haskell Code>biglistmean 10000000 5000000.5 C:\Documents and Settings\Travis\My Documents\Haskell Code>biglistmean 100000000 50000000.5 C:\Documents and Settings\Travis\My Documents\Haskell Code>biglistmean 1000000000 500000000.067109 On the final test of 10^9, Don reports that it took 1.76 secs on his machine. In contrast, just 10^8 takes 12.63 secs on my machine (sophisticatedly timed with handheld stopwatch) and on the coup de grace 10^9 test, it takes 2min:04secs. Yikes! My hardware is a little old (Win XP on Pentium 4 3.06GHz w 2 GB RAM) but not THAT old. I'm using the latest Haskell Platform which includes ghc v 6.10.4. Primary question: What gives here? Incidental questions: Is there a nice way to time executed code in Windows ala the "time" command Don shows under Linux? Also, does the ordering of the compiler flags have any impact (I hope not, but I don't want to be surprised ...) Thanks, Travis Erdman

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Translating a while
by Nicolas Couture-Grenier 08 Mar '10

08 Mar '10

I'm learning Haskell and I'm trying to translate a pseudocode algorithm to generate the next permutation from a previous permutation. A permutation is a list of n numbers (let's call it a) in {1 .. n} appearing once in arbitrary order. The first step is to find the largest index j in the list for which a[j] < a[j+1]. The pseudocode is simple: j:= n-1 while a[j] > a[j+1] j:=j-1 I've coded a haskell function to do this, but it is much uglier than the pseudocode : j :: Integral a => [a] -> Int j [] = 0 j xs = if (head (tail (reverse xs)) < last xs) then (length xs)-2 else j (take (length xs - 1) xs) Does anyone has a more elegant solution for this first step?

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Help in pattern matching
by Joe Fox 08 Mar '10

08 Mar '10

Hi Guys , I am having a problem with patter matching . I have declared following data type data TestData = DataConst1 Int String | DataConst2 Int Int | DataConst3 Sting String | DataConst4 String Int and now i have function *"test"* which behaves similarly for (DataConst1 and DataConst2 ) and behaves similarly from DataConst3 and DataConst4. for ex test (DataConst1 x y) = x test (DataConst2 x y )= x test (DataConst3 x y) = y test(DataConst4 x y )= y no my problem here is , is there any way i can merge these 4 statements in two statements .... some way i can match DataConst1 and DataConst2 in the same line it would save me a lot of rewriting of code

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Call for papers: MULTICONF-10, USA, July 2010
by John Edward 05 Mar '10

05 Mar '10

It would be highly appreciated if you could share this announcement with your colleagues, students and individuals whose research is in computer science, computer engineering, information science and related areas. Call for papers: MULTICONF-10, USA, July 2010 The 2010 multi-conference (MULTICONF-10) (website: http://www.PromoteResearch.org ) will be held during July 12-14, 2010 in Orlando, Florida, USA. The primary goal of MULTICONF is to promote research and developmental activities in computer science, information technology, control engineering, and related fields. Another goal is to promote the dissemination of research to a multidisciplinary audience and to facilitate communication among researchers, developers, practitioners in different fields. The following conferences are planned to be organized as part of MULTICONF-10. • International Conference on Artificial Intelligence and Pattern Recognition (AIPR-10) • International Conference on Automation, Robotics and Control Systems (ARCS-10) • International Conference on Bioinformatics, Computational Biology, Genomics and Chemoinformatics (BCBGC-10) • International Conference on Computer Communications and Networks (CCN-10) • International Conference on Enterprise Information Systems and Web Technologies (EISWT-10) • International Conference on High Performance Computing Systems (HPCS-10) • International Conference on Information Security and Privacy (ISP-10) • International Conference on Image and Video Processing and Computer Vision (IVPCV-10) • International Conference on Software Engineering Theory and Practice (SETP-10) • International Conference on Theoretical and Mathematical Foundations of Computer Science (TMFCS-10) MULTICONF-10 will be held at Imperial Swan Hotel and Suites. It is a full-service resort that puts you in the middle of the fun! Located 1/2 block south of the famed International Drive, the hotel is just minutes from great entertainment like Walt Disney World® Resort, Universal Studios and Sea World Orlando. Guests can enjoy free scheduled transportation to these theme parks, as well as spacious accommodations, outdoor pools and on-site dining — all situated on 10 tropically landscaped acres. Here, guests can experience a full-service resort with discount hotel pricing in Orlando. Please see the website http://www.PromoteResearch.org for more details. Sincerely John Edward

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Simple IO problem
by Ali Razavi 05 Mar '10

05 Mar '10

Why doesn't this work the way it's supposed to, or the way it's intuitively apparent from the code, that is, showing the prompt first, getting the line next, and printing the result finally? main = do putStr "Please Enter Your Name: " name <- getLine putStrLn ("Hello " ++ name) changing putStr with putStrLn rectifies it to the expected behavior, but I wonder why this version misbehaves. FWIW, I use ghc in cygwin. Ali

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