- Beginners - Haskell.org

very impure [global] counter
by Davi Santos 23 Jul '11

23 Jul '11

Hello all, I have massive (parallel if possible) system calls to an external non-deterministic program. Each time it is executed, it creates a file depending on a command line option 'opt' (input files path, for example). How can I ensure the file name will be unique? maybe with a global counter? My temporary solution have been to use a large random number: ----------- mysteriousExecution :: String -> IO () mysteriousExecution opt = do number <- rand run $ "mysterious-command " ⊕ opt ⊕ " --create-file=" ⊕ number rand = do a ← getStdRandom (randomR (1,999999999999999999999999999999999)) ∷ IO Int let r = take 20 $ randomRs ('a','z') (mkStdGen a) ∷ String return r ======== I'm trying to avoid additional parameters to 'mysteriousExecution'. I tried a counter also (to replace rand), but I don't know how could I start it inside 'mysteriousExecution'. c ∷ IO Counter c = do r ← newIORef 0 -- start return (do modifyIORef r (+1) readIORef r) If somebody says everything is wrong, ok. I understand. 18 years of imperative programming world can damage the brain. Thanks

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problem exercise 3 page 60 Programming in Haskell
by Roelof Wobben 22 Jul '11

22 Jul '11

Hello, I don't see the answer here. If I brake down the problem. We have this : [(x,y) | x <- [1..3], y <- [4,5,6]] I have to use one generator with two nested list compreshession and make use of concat. So the x generator is [x | x <- [1..3]] and the Y genarator is [y| y <- [4..6]] So if I put it together it uses the numbers [1..6] so I could do something like this in pseudo code as guard. If generator smaller or equal 3 then its x else it's a y. But if I do concat [x,y] then I get [1..6] and that not good. If I use zip [x,y] then I get [ (1,4) (2,5) (3,6)] which is also not good but better. So im stuck now and im puzzeling the whole afternoon about this ? Anyone who can give me a hint ? Roelof

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find problem
by Roelof Wobben 22 Jul '11

22 Jul '11

Hello, Sorry for asking so much but I really want to understand this. I have the function find k t [v, (k',v') <- t,k = v] Let's say k = false and t = [false, true, false, true] Then It will be find "false" ["false", "true", "false", "true] But where on earth are t,k = v and (k', v') come from ? Roelof

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Re: [Haskell-beginners] another list comprehesion error
by Thomas 22 Jul '11

22 Jul '11

On 22.07.2011 08:46, Roelof Wobben wrote: >> Now try this instead: >> roelof' n = [x | x<- [1..n]] >> What do you get for "roelof' 4" ? > A error message that a instance of print is missing. Then you have a typo somewhere... >> Now you can try the last two with guards. > 2) cannot be done without guards and list comprehession generator [ 2,4 ..10] does not work Sure it can: [ 2*x-1 | x <- [1..5]] and even easier: [1,3..10] or, better still (but not only a list comprehension any more): take 5 [1, 3..] > 3) cannot be done withut guards and list comprehession because of the a<=b This, too, can be done: [ (b, a) | a <- [1..5], b <- [1..a]] Regards, Thomas

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Haskell state monad example - type mismatch error
by Rohit Garg 22 Jul '11

22 Jul '11

Hi, I am trying out a simple haskell state monad example. I think I have understood the concept of monads, but I am getting stuck at using State monad. As far as I understand, the code below should compile, but it is throwing a type mismatch error in the argument to show. The rest of the code, however, type checks all right. If any one can point out what I am doing wrong, it would be really helpful. Thanks and regards, Rohit =============================== import Control.Monad.State import Data.Word type LCGState = Word32 lcg :: LCGState -> (Integer, LCGState) lcg s0 = (output, s1) where s1 = 1103515245 * s0 + 12345 output = fromIntegral s1 * 2^16 `div` 2^32 seed :: LCGState seed = 5 getRandom :: State LCGState Integer getRandom = do s0 <- get let (x,s1) = lcg s0 put s1 return x addThreeRandoms :: State LCGState Integer addThreeRandoms = do a <- getRandom b <- getRandom c <- getRandom return (a+b+c) main :: IO () main = putStrLn show(addThreeRandoms seed) -- Rohit Garg http://rpg-314.blogspot.com/

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FW: another list comprehesion error
by Roelof Wobben 22 Jul '11

22 Jul '11

---------------------------------------- > From: rwobben(a)hotmail.com > To: haskell(a)phirho.com > Subject: RE: [Haskell-beginners] another list comprehesion error > Date: Fri, 22 Jul 2011 06:46:03 +0000 > > > > > ---------------------------------------- > > Date: Thu, 21 Jul 2011 23:40:52 +0200 > > From: haskell(a)phirho.com > > To: beginners(a)haskell.org > > Subject: Re: [Haskell-beginners] another list comprehesion error > > > > Hi! > > > > On 21.07.2011 21:29, Roelof Wobben wrote: > > > roelof n x = [x | y<- [1..n]] > > > > Have you tried this? > > What do you get for, say "roelof 4 5" ? Why? I get [5,5,5,,5] And that the right answer according to the exercise > > Now try this instead: > > roelof' n = [x | x <- [1..n]] > > What do you get for "roelof' 4" ? A error message that a instance of print is missing. > > A few suggestions (without guards): > > The list of the first 5 natural numbers. > > The list of the first 5 odd numbers. > > The list of all pairs (a, b) where a,b > 0, a <= b and b <= 5 > > > > Now you can try the last two with guards. without guards 1) [x | x <- [1..5]] 2) cannot be done without guards and list comprehession generator [ 2,4 ..10] does not work 3) cannot be done withut guards and list comprehession because of the a<=b with guards. 1) has no need for guards. 2) [x | x <-[1..10], even x] 3] [ (a,b) | a<- [1..4], b<- [1..5], a<=b] Roelof

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another list comprehesion error
by Roelof Wobben 21 Jul '11

21 Jul '11

hello, I know try to make this work. The user enters a number and the programm calculates all the numbers which are smaller then the number and where x^2+Y^2=Z^2. So i Thought this would work. roelof :: a -> b -> c -> (a,b,c) roelof n = [(x y z) | x^2+Y^2=Z^2 <- x<-[1..n], y<- [1..n], z<-[1..n]] But I get this error : oefening.hs:2:30: parse error on input `=' Roelof

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Re: [Haskell-beginners] Typeclasses vs. Data
by Thomas 21 Jul '11

21 Jul '11

Hello David! Yes, I should have posted the signatures, too. Sorry. I managed to simplify the problem even further. It still doesn't help me understand how to avoid the error, though. class Continuation a where resume :: a -> Int -> Int data BeginCont a = BeginCont a Int deriving (Show) instance (Continuation a) => Continuation (BeginCont a) where resume (BeginCont k es) v = eval_begin es k eval :: Continuation a => Int -> a -> Int eval n k = if n < 1 then resume k n else eval_begin (n - 1) k eval_begin :: Continuation a => Int -> a -> Int eval_begin n k = eval n (if (n < 0) then k else (BeginCont k (n - 1))) It's pretty clear that in the mutual recursion between "eval" and "eval_begin" the parameter "k" is "growing" from "a" to "BeginCont a" and so on. But how to resolve that? (And it works perfectly in the case of the "data" definition.) Anyway, thank you! Regards, Thomas On 21.07.2011 00:40, David Place wrote: > On Jul 20, 2011, at 6:26 PM, Thomas wrote: > >> Thank you for taking the time. >> Here is a complete fragment that shows the error: > > Hi, Thomas. > > I'm very sympathetic. I hate it when I get an error like this. I looked at your code and the solution didn't jump off the page, maybe it will for someone else. In the meantime, I suggest this strategy. Carefully give type signatures to all of your functions. This way you can help the type checker give better error messages. The type inference algorithm can go away into crazy land if you give it a nonsense definition. > > ___________________ > David Place > Owner, Panpipes Ho! LLC > http://panpipesho.com > d(a)vidplace.com > >

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Drawing Information from a function already defined
by Clockwork PC 21 Jul '11

21 Jul '11

Greetings Haskell community, This is my first ever post so please be gentle... Here's where I got to: Jumped into GHCi: GHCi, version 7.0.3: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer-gmp ... linking ... done. Loading package base ... linking ... done. Defined my function: Prelude> let rightTriangles = [ (a,b,c) | c <- [1..10], b <- [1..10], a <- [1..10], a^2 + b^2 == c^2 ] Tested my function: Prelude> rightTriangles [(4,3,5),(3,4,5),(8,6,10),(6,8,10)] Now, I want to define a refinement of this that will only select values whose total is 24. I could write it so: let rightTriangles = [ (a,b,c) | c <- [1..10], b <- [1..10], a <- [1..10], a^2 + b^2 == c^2,* a+b+c == 24*] However, it seems like a bit of a waste if I already have "rightTriangles" defined. I tried a few things, but none of them worked: Prelude> let rightTriangles` = rightTriangles, a+b+c == 24 <interactive>:1:21: parse error on input `=' Prelude> let rightTriangles` = rightTriangles | a+b+c == 24 <interactive>:1:21: parse error on input `=' Prelude> let rightTriangles` = [rightTriangles, a+b+c == 24] <interactive>:1:21: parse error on input `=' Prelude> let rightTriangles` = [rightTriangles | a+b+c == 24] <interactive>:1:21: parse error on input `=' Basically, I'm trying to work out how to draw data from a list already to hand. Alexander

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Re: [Haskell-beginners] What book is being referred to: "... in SPJs beautiful code chapter ..."
by Erlend Hamberg 21 Jul '11

21 Jul '11

On 21 Jul 2011 13:34, "Costello, Roger L." <costello(a)mitre.org> wrote: > Do you know what book is being referred to? I'm quite sure it's “Beautiful Code”: http://oreilly.com/catalog/9780596510046 -- Erlend Hamberg ehamberg(a)gmail.com

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