May 2012 - Beginners - Haskell.org

Missing some functions in Hoogle
by Ken Kawamoto 26 May '12

26 May '12

Hi, I'm looking for a way to search Hoogle for Yesod related functions, say, renderRoute in Yesod.Routes.Class. First I confirmed the Hoogle site doesn't have it in index. http://www.haskell.org/hoogle/?hoogle=renderRoute Then, after installing hoogle via cabal, I built index over all packages that I have by "hoogle data -l -v all". Below is an excerpt of the output, looks like yesod-routes is imported to database, or indexed, correctly. However, local Hoogle still doesn't return any entry when I searched for "renderRoute". ---- "hoogle data -l -v all" output Starting yesod Starting yesod-auth Starting yesod-core Starting yesod-routes Skipped 1 warnings in yesod-routes Parse error 11:48: Parse error: family Converting yesod-routes... done Finished yesod-routes Skipped 4 warnings in yesod-core Parse error 683:77: Parse error in expression: _ Parse error 710:48: Parse error: family Parse error 803:1: Can't translate Parse error 806:1: Can't translate Converting yesod-core... done Finished yesod-core Now I'm wondering if I'm missing something, or Hoogle supports only a predefined set of functions. Any advise would be appreciated. -- Ken

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Re: [Haskell-beginners] Stymied by mutable arrays in the ST monad
by Matthew Moppett 25 May '12

25 May '12

Brilliant. I think I've got it now. Thanks heaps, Tobias. On Sat, May 26, 2012 at 4:01 AM, Tobias Brandt <tob.brandt(a)googlemail.com>wrote: > On 25 May 2012 19:43, Matthew Moppett <matthewmoppett(a)gmail.com> wrote: > > About the extra type info needed -- what part of the type "ST s (STArray > s > > Int Int)" is the compiler unable to infer? > > It doesn't know which array type to use, because 'newListArray' is > polymorphic > in the type of array it returns. This type is not mentioned in > the signature of idST because the array is immediately consumed again by > 'getElems'. > > You can try it out in ghci (':t expr' gives you the type of an expression): > > :m +Data.Array.ST > let xs = [1,2,3 :: Int] > > :t xs > [Int] > > :t newListArray (1, length xs) xs > MArray a Int m => m (a Int Int) > > Applying getElems gets you this: > :t (newListArray (1, length xs) xs) >>= getElems > MArray a Int m => m [Int] > > From the 'runST' in your program, the compiler knows that m = ST s, > for some s, but there is nothing constraining 'a' further. > Any 'a' fulfilling the constraint 'MArray a Int (ST s)' will do. > Therefore we have to fix 'a' to some type, e.g. 'STArray s', but > 'STUArray s' would work too, in this case. >

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Stymied by mutable arrays in the ST monad
by Matthew Moppett 25 May '12

25 May '12

I've been trying to use mutable arrays in the ST monad, and wrote out a little proof of concept function: idST :: [Int] -> [Int] idST xs = runST $ do array <- newListArray (1, (length xs)) xs return (getElems array) -- where idSt should be equivalent to id. And I get the error message: Couldn't match type `[Int]' with `Int' In the return type of a call of `getElems' In the first argument of `return', namely `(getElems array)' In a stmt of a 'do' block: return (getElems array) Obviously I'm making a very simple mistake here, but I can't seem to spot it. Can anyone offer some advice?

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Help!, was Re: Trouble with "import qualified"
by Dudley Brooks 24 May '12

24 May '12

Hi. I've had this up for over a week, and haven't received any response. Any help would be greatly appreciated. On 5/16/12 10:04 PM, Dudley Brooks wrote: > Almost complete beginner. > > I just downloaded Haskell 32-bit for Mac from Haskell.org and am going > through Learn You a Haskell. I tried to follow the instructions on > > http://learnyouahaskell.com/modules#loading-modules > > in particular > > import qualified Data.Map as Map > > which, following the instructions, I put in a .hs file which I attempted > to load into GHCi with :l. I got > > parse error on input `import' > > I have no idea what the problem might be or what to do about it. Other > imports have been successful, but after getting the above message I > tried to "import qualified" other modules, always with the same error.

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Traverse tree with computing current level using Foldable instance.
by Dmitriy Matrosov 24 May '12

24 May '12

Hi. I can't figure out how should i properly solve the following problem. There is a tree defined like data Tape a = Tape a [Tape a] and i want to traverse it in some specific order, computing at the same time current level (depth). I.e. it should like fold, and folding function should have access to current level in the tree. Here is my implementation: import Data.Monoid import Control.Monad.State type TapeState a = State Int a foldMapS2 :: (Monoid m) => (a -> TapeState m) -> TapeState (Tape a) -> TapeState m foldMapS2 f tt = do t@(Tape name ts) <- tt foldr (go f) (f name) ts where go :: (Monoid m) => (a -> TapeState m) -> Tape a -> TapeState m -> TapeState m go f t mz = do cs <- get x <- foldMapS2 f (State (\s -> (t, s + 1))) put cs z <- mz put cs return (x `mappend` z) and here is example usage testTape = Tape "A" [ Tape "B" [ Tape "C" [] , Tape "F" [Tape "G" [Tape "H" []]] , Tape "E" [] ] , Tape "D" [ Tape "I" []] ] *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 0 then return [name] else return mempty) (return (testTape))) 0 (["A"],0) *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 1 then return [name] else return mempty) (return (testTape))) 0 (["B","D"],0) *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 2 then return [name] else return mempty) (return (testTape))) 0 (["C","F","E","I"],0) *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 3 then return [name] else return mempty) (return (testTape))) 0 (["G"],0) *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 4 then return [name] else return mempty) (return (testTape))) 0 (["H"],0) *Main> runState (foldMapS2 (\name -> get >>= \cs -> if cs == 5 then return [name] else return mempty) (return (testTape))) 0 ([],0) As you can see, this just selects all elements at particular tree level. So, my foldMapS2 looks similar to foldMap from Foldable, but i can't figure out, how should i define instances of Foldable (and Monoid?) to achieve the same functionality?

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Most C++ compilers will not optimize x^2.0 as x*x but instead will do an expensive ...
by KC 24 May '12

24 May '12

exponentiation and logarithm. So, I believe this C++ versus Haskell versus (your language of choice) is a Penn & Teller misdirection. Whereas, another level of indirection solves everything. -- -- Regards, KC

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Found quote: "One of the ineradicable illusions about C++ is that the compiler should be able to see through ...
by KC 23 May '12

23 May '12

all the abstractions and obfuscations an “advanced” C++ program contains. First and foremost, C++ should be seen as a language that enables complexity management. The features one has grown fond of in this concept, like operator overloading, object orientation, automatic construction/destruction, etc., are however mostly unsuitable for efficient low-level code." >From "Introduction to High Performance Computing for Scientists and Engineers" CRC Press, 2011. So if one avoids such C++ complexity management features and a Haskell compiler can see through most of the complexity management features of Haskell; they should be comparable. :) Casey -- -- Regards, KC

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From some "High-Performance Computing" book; "It is hard to eradicate from C++'ers the need for objects, classes, etc. ..."
by KC 22 May '12

22 May '12

"For high-performance computing you want to avoid such software engineering abstractions as classes, objects, etc...." Therefore, I would say, since Haskell makes possible many fine-levels of abstraction, if the compiler can strip most of these away, then it would come close to the speed of C (i.e. C++ without most of the software engineering abstractions). Still, I remember that quote that, "Another level of indirection solves everything." Or If you're Penn & Teller, "Another level of misdirection solves everything." Casey -- -- Regards, KC

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[x] and (x:_) for lists -- did you ever think that odd?
by AntC 22 May '12

22 May '12

Think back to when you first came across Haskell ... This, for example (from page 4 of the Gentle Intro -- my comments added): length :: [a] -> Integer -- [ ] means list length [] = 0 -- [] means list length (x: xs) = 1 + length xs -- list, but no [ ] Usually, showing a list uses square brackets and comma separators. So list literals use square brackets and comma separators. Type decls for a list uses square brackets. List builders use square brackets and commas -- such as [1, 3 .. 9] or [0 ..] Pattern matching for finite length lists use square brackets and commas: f [] = ... f [x] = ... f [x, y] = ... But pattern matching for unknown-length lists uses round brackets and colon -- such as that last binding for `length` above. And (nearly) every list-handling function has a pattern for unknown-length lists. Would this pattern matching seem less odd?: head [x ..] = x length [x, xs@..] = 1 + length xs Experienced Haskellers need not answer: you've got too used to ( : ) ;-) AntC

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haskell for system administration
by Ben 21 May '12

21 May '12

Hi, I am new to HASKELL.Currently I am using shell scripting for Administration tasks and all.Kindly suggest me , does HASKELL usefull for Linux system administration and reporting purpose.? Do we use HASKELL for network programming and all? Regards, Ben

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