- Beginners - Haskell.org

New List
by Chad Wilson 28 Jul '08

28 Jul '08

And the new list is now begun for people like me, n00bs. -- Chad Wilson

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[Haskell-begin] Cabal-Install Update Problems
by Jeff Wheeler 26 Jul '08

26 Jul '08

Hi, (Note: I also posted this a few days ago to libraries(a)haskell.org without much response, and thought I might have more success here. Please correct me if this is inappropriate.) I'm trying to use cabal-install on two machines, Debian Sid and OS X 10.5, but I am unable to run "update" on either. Most other commands seem to work fine, but the networking for "update" gives an error on both machines, completely independently: Downloading package list from server 'http://hackage.haskell.org/packages/archive' cabal: recv: resource vanished (Connection reset by peer) Running with the -v3 option does not provide much more information: Downloading package list from server 'http://hackage.haskell.org/packages/archive' Sending: GET http://hackage.haskell.org/packages/archive/00-index.tar.gz HTTP/1.1 User-Agent: cabal-install/0.5.1 Creating new connection to hackage.haskell.org cabal: recv: resource vanished (Connection reset by peer) Both machines are using a vanilla Netgear ProSafe VPN Firewall FVS318v3, a very simple prosumer router. Obviously, the router works fine for everything else, and has been running flawlessly for ages. This is using cabal-install from darcs and the cabal 1.4 branch. Thanks a bunch for helping, Jeff Wheeler

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[Haskell-begin] Morphing Endo (ICFP Contest 2007)
by Rafael Gustavo da Cunha Pereira Pinto 25 Jul '08

25 Jul '08

Did any of you tried to do a Haskell implementation of the ICFPC 2007 problem? I was thinking of using it as a learning exercise, but I am afraid of the stack. My approach is: 1) use Data.ByteString.Lazy.Char8 to read the contents of the DNA file 2) create a recursive function process::ByteString -> a that will call itself. I have a few problems: a) the DNA is 8MB long. How can I ensure the stack will hold a recursive call? b) there is an "abnormal ending" function called finish that is called anywhere in the code. Is it a good approach to return Empty to end processing? c) should I go "monadic", keeping the dna on a state monad? Thanks -- Rafael Gustavo da Cunha Pereira Pinto Electronic Engineer, MSc.

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[Haskell-begin] some basic syntax questions
by Anatoly Vorobey 25 Jul '08

25 Jul '08

Hello Haskell-beginners, I'm trying to work through the "Write Yourself a Scheme in 48 Hours" tutorial. One of the first exercises calls for writing code to sum up arguments on command line and display the sum. After a few tries, this worked for me: module Main where import System.Environment main :: IO () main = do args <- getArgs putStrLn ("Hello, " ++ show (sumIt args)) where sumIt x = sum $ map read x However, I have two basic questions: 1. I initially tried putStrLn("Hello, " ++ show $ sumIt args), but that didn't compile. Why not? If I wrap (show $ sumIt args) in parens, it works, but should I have to, if I didn't have to in the original code above? 2. I initially tried where sumIt = sum $ map read (the "point-free" style, I believe it's called?) but that didn't compile. Why not? A friend suggested where sumIt = sum . map read and that does work; I guess my real problem, then, is that I don't really understand the difference between the two and why the former doesn't work. Thanks in advance, Anatoly. -- Anatoly Vorobey, avorobey(a)gmail.com http://avva.livejournal.com (Russian) http://www.lovestwell.org (English)

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Re: [Haskell-begin] Morphing Endo (ICFP Contest 2007)
by Rafael Gustavo da Cunha Pereira Pinto 24 Jul '08

24 Jul '08

Ops, forgot to copy the list. On Thu, Jul 24, 2008 at 07:54, Rafael Gustavo da Cunha Pereira Pinto < rafaelgcpp(a)gmail.com> wrote: > > > > > 2008/7/23 Benjamin L. Russell <DekuDekuplex(a)yahoo.com>: > >> >> Correct me if I'm wrong, but because the prefix should not be too >> long, and chemical process of modifying Endo with the prefix should >> not consume too much energy, this appears to be an example of >> constraint programming. Since constraint programming can often be >> carried out by constraint logic programming, I might suggest a logic >> programming approach for this problem. >> >> > > The DNA is actually a representation of a turing machine (like BF > language). Using brute-force and constraints would take too long. There is a > report at http://www.cs.uu.nl/research/techreps/repo/CS-2007/2007-029.pdf, > that shows how this DNA string was made: it is the output of a compiler for > a functional programming language specially tailored for this contest. > which, in turn, was written in Haskell. > > The basic idea is to disassemble, find the function calls and add > prefxes that do the appropriate calls. > > > -- > Rafael Gustavo da Cunha Pereira Pinto > Electronic Engineer, MSc. > -- Rafael Gustavo da Cunha Pereira Pinto Electronic Engineer, MSc.

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[Haskell-begin] data declarations
by Roger Myers 23 Jul '08

23 Jul '08

hi Not sure if I'm on the right track here, so as a beginner I thought I'd ask on this list. I want to declare a type 'All' that consists of the of the types 'Tinteger, Tbool, Tstring, Tarray' but I would also like to make a relationship that Tinteger and Tbool are also the type 'Simple' and that 'Tstring' and 'Tarray' are 'Composite' and that Mixed is then either Simpletype or Composite. So I can then write a function that takes a type of All and produces a list of Mixed. data All = Tinteger Int | Tbool Bool | Tstring String | Tarray [Alltypes] data Simple = Tinteger | Tbool data Composite = Tstring | Tarray data Mixed = Simple | Composite k :: All -> [Mixed] k a = case a of Simple s -> [Simple s] Composite c -> [Composite c] But this isn't valid Haskell code because I can't have multiple declarations of the same types, is there a way to specify a relationship, if so can someone point me in the right direction? Roger

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[Haskell-begin] Exercises for beginners and Natural Tansformations
by Federico Brubacher 22 Jul '08

22 Jul '08

Hi all, I think i will love this list :) Yesterday I was looking at this Haskell Exercises for beginners [1] and got somewhat stuck at Ex nbr 2 , here's why: The idea of this exercises is that you don't use built in Haskell Lists, so no pattern matching, instead you have to use the data type list provided in the source of the exercise. My friend who is more advanced in Haskell said this approach is good because you get to learn natural Transformations. So I googled that and I came up with this [2]. I can read al little bit of category theory but I would love to know how to apply [2] in the sum exercise. Anyway as we all know summing all the elements in a list using Haskell lists is trivial : sum :: [Int] -> Int sum (xs) = foldr(+)0xs But how to do it with Natural transformations ??? Thanks a lot Federico [1] http://blog.tmorris.net/haskell-exercises-for-beginners/ -- Federico Brubacher www.fbrubacher.com Colonial Duty Free Shop www.colonial.com.uy

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[Haskell-begin] Using read, reads...
by Rafael Gustavo da Cunha Pereira Pinto 21 Jul '08

21 Jul '08

Hi folks, I am trying to make a program that reads a file which lines contain a series of 6 integers. -------------sample.in------------- 01 02 03 04 05 06 10 11 15 18 29 45 19 22 10 01 23 14 ----------------------------------------- right now I am using hGetLine and getting a string. Is is possible to use read to convert this in a list of Num? -- Rafael Gustavo da Cunha Pereira Pinto Electronic Engineer, MSc.

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[Haskell-begin] Re: [Haskell-cafe] Trouble with non-exhaustive patterns
by C.M.Brown 21 Jul '08

21 Jul '08

Hi Fernando, I hope you don't mind, but I've moved this over to the Haskell-beginners mailing list, where I think this kind of question will be more appropriate. In Haskell, it helps to think of functions in terms of an input and an output, that is, what is the thing that is going into the function; and what is the thing that is coming out of the function? In your function, final, the input is clearly a list of something (we can denote this by the Haskell type [a] which means a list of some type, a). Its return type is clearly an element of the list, so that must be the something type (the 'a' from the [a]). This gives the Haskell type: final :: [a] -> a (final takes a list of 'a' and gives back a single 'a'. The 'a' is a type variable, and it is used to denote that anything can be put in its place, so we can give final a list of integers, characters, whatever). Now, let's take a look at your definition of final. If we take a closer look, in fact only two equations satisfy this type: final [a] = a final (_:t) = final t The other takes a list and returns a list. The equation, final [] = [] takes an empty list and returns an empty list (its type is therefore [a] -> [a]). This is why you got an error, as Haskell doesn't know how what to do with the conflicting equation. What we need is the final element of the list. How do we do that? Let's think of the simple cases first. The final element of a list containing a single element is just that element, final [a] = a But what about if the list contains more elements? Or is an empty list? The empty list may be confusing, as an empty list contains no elements, so in effect, we can't return anything. We can, however, return an error message. fun [] = error "empty List" And the final element of any list, must be the final element of its tail: final (_:t) = final t this gives us: final :: [a] -> a final [] = error "Empty List" final [a] = a final (_:t) = final t I hope that gives some insight. Kind regards, Chris. On Mon, 21 Jul 2008, Fernando Rodriguez wrote: > > Hi, > > I defiend the following function to get the last element of a list: > > final [a] = a > final (_:t) = final t > > and it works as expected. Since I didn't want to have a non exhaustive pattern, > I added the following case: > > final [] = [] - I consider that the end of an empty list is the empty list > final [a] = a > final (_:t) = final t > > Suddenly, the function stoped working with a rather cryptic (for a newbie > at least) error message: > > *Temp> final [4,5] > > <interactive>:1:9: > No instance for (Num [a]) > arising from the literal `5' at <interactive>:1:9 > Possible fix: add an instance declaration for (Num [a]) > In the expr*Temp> ession: 5 > In the first argument of `final', namely `[4, 5]' > In the expression: final [4, 5] > > What have I done so wrong? > > Thanks in advance, > Fernando > > > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe(a)haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe >

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[Haskell-begin] Alternating sequence
by Dirk Markert 21 Jul '08

21 Jul '08

I am trying to generate the following list: 2, 3, 5 -- and then alternating (+2) resp (+4) -- 7, 11, 13, 17, 19, 23 I came up with the following solution 2:3:unfoldr (\(a,b) -> Just (a,(a+b, if b == 2 then 4 else 2))) (5,2) Are there easier ways to generate the desired list? Dirk

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