- Beginners - Haskell.org

safetail problem
by Roelof Wobben 13 Jul '11

13 Jul '11

Hello, As a exercise I need to rewrite the tail function into a safetail function by using conditions. So I made this : http://codepad.org/bKcCUdqy But as you can see there is a error message. Can anyone explain me how to solve this ? Roelof

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Problems with typeclasses instances
by ARJANEN Loïc Jean David 13 Jul '11

13 Jul '11

I have the following problem with typeclasses : I have created a type class in a manner similar to the following module Dummy where import Char class (Show t) => Dummy t where kite :: t -> t instance (Dummy t) => Dummy [t] where kite = reverse instance Dummy String where kite = map toUpper And use it thus : import Dummy main :: IO () main = do print.kite $ "Rest in peace" If I compile without any option, I get as expected an error because of the type synonym instance. If I remove the instance for String, I also get as expected an error because Char is not an instance of Dummy. But if I add (as suggested in the first error message) the GHC's option -XTypeSynonymInstances, I get an error because of overlapping instances, even if Char is still not an instance of Dummy and I don't see how String could be eligible for the (Dummy t) => Dummy [t] instance... Thanks in advance for your help, ARJANEN Loic.

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User-defined polymorphic data type: heterogeneous list?
by Christopher Howard 13 Jul '11

13 Jul '11

I'm trying to understand parametric polymorphism and polymorphic data types. I especially want to go beyond simply using the defined polymorphic data types (lists and so forth) and see how I can make my own useful ones. As my first stab at it, it seemed like I should be able to create my own heterogeneous "list" data type -- i.e., a "list" data type that can contain elements of different types. (like, [3,'a',True], for example) But I'm a little stuck. My first try was like so: data HeteroList a b = Null | Element a (HeteroList a b) deriving (Show) ...but this of course did not work, because all elements end up having to be the same type. Then I tried data HeteroList a b = Null | Element a (HeteroList b a) deriving (Show) ...but this doesn't work because every other other element has to be the same type: Element 'a' (Element 1 (Element 'a' (Element 2 Null))) ...I could go on and embarrass myself some more, but since I'm likely widely off-base I'll just ask if somebody can point me in the right direction. -- frigidcode.com theologia.indicium.us

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(Integral a) => a vs Integer
by Dan Ross 12 Jul '11

12 Jul '11

Hi all- I'm going through Learn You a Haskell for Great Good and I don't understand the advantage of using (Integral a)=> a vs just Integer as I show below. Could someone explain this to me? Thanks, Dan lucky :: (Integral a) => a -> String lucky 7 = "LUCKY NUMBER SEVEN!" lucky x = "Sorry, you're out of luck, pal!" lucky :: Integer -> String lucky 7 = "LUCKY NUMBER SEVEN!" lucky x = "Sorry, you're out of luck, pal!"

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Re: [Haskell-beginners] function defenition. Do I understand it right?
by Roelof Wobben 12 Jul '11

12 Jul '11

Hello, I had a look at it but I don't think it will solve my problem. What I try to achieve is this: [1,2,3,4] will be [1,2] [3,4] [1,2,3,4,5] will be [1,2,3] [3,4,5] So the halfs have to be the same lenght. So I use : length xs'div' 2 to figure out where the middle is. lenght xs'mod'2 to look if it's a even or not even list. So I can use splitat but then the lists will not be the same length if it's a not even list. Roelof ---------------------------------------- > Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? > From: jhenahan(a)uvm.edu > Date: Mon, 11 Jul 2011 15:55:13 -0400 > To: rwobben(a)hotmail.com > > You're overcomplicating things. Look at splitAt from Data.List: > > http://hackage.haskell.org/packages/archive/base/4.1.0.0/doc/html/Data-List… > > On Jul 11, 2011, at 3:51 PM, Roelof Wobben wrote: > > > Hello, > > > > Thanks for the tip. > > Im stuck now. > > > > I have this : > > > > halve (xs) | length xs'mod'2 == 0 = (take n xs drop n xs) > > | otherwise = (take n+1 xs drop n+1 xs > > where n= lenght xs'div'2 > > > > > > main = do > > putStrLn $ show $ halve [1,2,3,4] > > putStrLn $ show $ halve [1,2,3] > > > > But now I get this message : > > > > ERROR line 3 - Syntax error in expression (unexpected keyword "where") > > > > Appearently I can't use where here. > > > > How can I solve this ? > > > > Roelof > > > > > > > > > Date: Mon, 11 Jul 2011 11:44:28 -0400 > > > Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? > > > From: dmcbride(a)neondsl.com > > > To: rwobben(a)hotmail.com > > > CC: beginners(a)haskell.org > > > > > > For simple snippets like this, you can go to > > > http://codepad.org/ddySVOPr and run your code to see if it works. > > > > > > And your first version (once corrected for syntax) works on any list > > > length except for empty lists. > > > > > > On Mon, Jul 11, 2011 at 10:48 AM, Roelof Wobben <rwobben(a)hotmail.com> wrote: > > > > Hello, > > > > > > > > I have made a exercise where I must split a even list. > > > > The answer is : > > > > > > > > halve xs = (take n xs drop n xs) > > > > where n = length xs 'div' 2 > > > > > > > > Now I want to change it so i can split even and not even list,. > > > > I thought this can be the answer > > > > > > > > halve xs = > > > > lenght xs 'mod' 2 = 0 : (take n xs drop n xs) > > > > otherwise : (take n+1 xs drop n+1 xs) > > > > where > > > > n = length xs 'div' 2 > > > > > > > > > > > > Can this be working ? > > > > I did not check this on GCHI because Im re - installing my OS. > > > > > > > > Regards. > > > > > > > > Roelof > > > > > > > > > > > > _______________________________________________ > > > > Beginners mailing list > > > > Beginners(a)haskell.org > > > > http://www.haskell.org/mailman/listinfo/beginners > > > > > > > > > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org > > http://www.haskell.org/mailman/listinfo/beginners >

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function defenition. Do I understand it right?
by Roelof Wobben 12 Jul '11

12 Jul '11

Hello, I have made a exercise where I must split a even list. The answer is : halve xs = (take n xs drop n xs) where n = length xs 'div' 2 Now I want to change it so i can split even and not even list,. I thought this can be the answer halve xs = lenght xs 'mod' 2 = 0 : (take n xs drop n xs) otherwise : (take n+1 xs drop n+1 xs) where n = length xs 'div' 2 Can this be working ? I did not check this on GCHI because Im re - installing my OS. Regards. Roelof

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Working out the types
by Patrick LeBoutillier 11 Jul '11

11 Jul '11

Hi all, Today after reading a question on StackOverflow I found myself "working out the types" for this expression: map . (:) Using a pencil and paper (and more time than I has hoped) I was able to come to the same answer as given by ghci: map . (:) :: a -> [[a]] -> [[a]] Here's my question: Does ghci have a verbose mode or something where is can show you step by step how the types are worked out? If not is there a hackage (or any other kind of) package that can do that? I've seen long examples worked out by people on this list (namely Daniel Ficher) and they are easy to follow and help me a lot, so I was wondering if such a program existed that could do it automatically. Thanks, Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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Re: [Haskell-beginners] function defenition. Do I understand it right?
by Jack Henahan 11 Jul '11

11 Jul '11

Forgot to CC the list. Begin forwarded message: > From: Jack Henahan <jhenahan(a)uvm.edu> > Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? > Date: July 11, 2011 3:55:13 PM EDT > To: Roelof Wobben <rwobben(a)hotmail.com> > > You're overcomplicating things. Look at splitAt from Data.List: > > http://hackage.haskell.org/packages/archive/base/4.1.0.0/doc/html/Data-List… > > On Jul 11, 2011, at 3:51 PM, Roelof Wobben wrote: > >> Hello, >> >> Thanks for the tip. >> Im stuck now. >> >> I have this : >> >> halve (xs) | length xs'mod'2 == 0 = (take n xs drop n xs) >> | otherwise = (take n+1 xs drop n+1 xs >> where n= lenght xs'div'2 >> >> >> main = do >> putStrLn $ show $ halve [1,2,3,4] >> putStrLn $ show $ halve [1,2,3] >> >> But now I get this message : >> >> ERROR line 3 - Syntax error in expression (unexpected keyword "where") >> >> Appearently I can't use where here. >> >> How can I solve this ? >> >> Roelof >> >> >> >>> Date: Mon, 11 Jul 2011 11:44:28 -0400 >>> Subject: Re: [Haskell-beginners] function defenition. Do I understand it right? >>> From: dmcbride(a)neondsl.com >>> To: rwobben(a)hotmail.com >>> CC: beginners(a)haskell.org >>> >>> For simple snippets like this, you can go to >>> http://codepad.org/ddySVOPr and run your code to see if it works. >>> >>> And your first version (once corrected for syntax) works on any list >>> length except for empty lists. >>> >>> On Mon, Jul 11, 2011 at 10:48 AM, Roelof Wobben <rwobben(a)hotmail.com> wrote: >>>> Hello, >>>> >>>> I have made a exercise where I must split a even list. >>>> The answer is : >>>> >>>> halve xs = (take n xs drop n xs) >>>> where n = length xs 'div' 2 >>>> >>>> Now I want to change it so i can split even and not even list,. >>>> I thought this can be the answer >>>> >>>> halve xs = >>>> lenght xs 'mod' 2 = 0 : (take n xs drop n xs) >>>> otherwise : (take n+1 xs drop n+1 xs) >>>> where >>>> n = length xs 'div' 2 >>>> >>>> >>>> Can this be working ? >>>> I did not check this on GCHI because Im re - installing my OS. >>>> >>>> Regards. >>>> >>>> Roelof >>>> >>>> >>>> _______________________________________________ >>>> Beginners mailing list >>>> Beginners(a)haskell.org >>>> http://www.haskell.org/mailman/listinfo/beginners >>>> >>>> >> _______________________________________________ >> Beginners mailing list >> Beginners(a)haskell.org >> http://www.haskell.org/mailman/listinfo/beginners >

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GHCI in my app...
by Sean Charles 10 Jul '11

10 Jul '11

Hi, I've now got a working OpenGL test-bed and I have also written myself a "GLI" as I call it, openGl command Line Interface... a forkIO thread that listens on port 3000 and once my program is up, I telnet in and issue commands to it so I can fiddle with OpenGL whilst it is running. I can change the camera position and angle and issue a few simple commands to affect the rendered environment so I can better understand how OpenGL works. If it works well and the code isn't to amateurish I will github it I guess! However, it seems to me that I am re-inventing the wheel to some degree: I am opening a socket, getting input, parsing it and executing commands.... what I would really like to be able to do is to *use* the already pretty slick ghci interface instead... ...so, is there a way to create an instance of it and attach it to the socket? f I run my app from within ghci it still works but the event loop steals any further input. So, is it possible. I *did* manage this with SLIME and Emacs some years ago albeit the slime funtionality was a little broken but I could actually recompile code and have it affect the still running application. I've also been using Parsec latelt and the "wai-handler-devel" also entered my mind as an option but I only want a simple CLI after all! :) Thanks for your upcoming suggestions..... Sean

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Re: [Haskell-beginners] Can i define a record without defining access method.
by yi huang 09 Jul '11

09 Jul '11

On Sat, Jul 9, 2011 at 6:45 PM, Tom Murphy <amindfv(a)gmail.com> wrote: > On 7/9/11, yi huang <yi.codeplayer(a)gmail.com> wrote: > > I'm trying to create a haskell implementation of json rpc, I try to > define > > protocol using record like this: > > > > data Request = Request { > > version :: Integer > > , id :: Integer > > , method :: String > > , args :: [Value] > > } deriving (Typeable, Data, Show) > > > > data Response = Response { > > version :: Integer > > , id :: Integer > > , code :: Integer > > , method :: String > > , result :: Value > > } deriving (Typeable, Data, Show) > > > > so i can use json library to encode/decode it. > > But this code fails, because haskell will define access function > > automaticlly, and function names conflicts. > > My question is, is there a way i can define record without access > function, > > so i can have same attribute name in multiple record. > > > If you don't want access functions defined, you can simply not name > >> your record fields: > >> > data Request = Request Integer Integer String [Value] > >> deriving (Typeable, Data, Show) > >> > > If you want it to be more readable, you can define type synonyms: > >> > type ID = Integer > >> type Version = Integer > >> [...] > >> > data Request = Request Version ID Method Args > >> deriving (Typeable, Data, Show) > >> > The two "instances" of ID won't conflict, then. > I have to define it as a record, so aeson can inspect the attribute names and encode it to a json object automatically, for example: "{\"args\":[],\"id\":1,\"method\":\"test\",\"version\":1}" But normal data constructor would be encoded to an array. Though i can define it as a normal data constructor, and implement ToJSON class manually, but the code would be more verbose. > > > Tom > -- http://www.yi-programmer.com/blog/

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