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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 23 May '12

23 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 21 May '12

21 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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Choosing a function randomly...
by Stuart Hungerford 20 May '12

20 May '12

Hi, This is kind-of related to my earlier question on looking up functions by name. Suppose I have a module with a number of functions with the same signature: scale :: Int -> Int -> Int scale s x = s * x add :: Int -> Int -> Int add a x = a + x ... I'd like to choose and run one of these functions randomly at run time. I can see I could use some kind of case expression: op :: Int -> Int -> Int op p x = case random(1,2) of 1 -> scale p x 2 -> add p x Or some kind of pattern guards: op p x | random(1,2) == 1 = scale p x | otherwise = add p x Although that method won't work as is for more than two choices. Are these methods the most idiomatic way of randomly choosing a function? How hard would it be to use the machinery of the QuickCheck library for this, given it must be doing something similar in test suites? Thanks, Stu

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Choosing a function by (string) name...
by Stuart Hungerford 20 May '12

20 May '12

Hi, Suppose I have a module of functions with identical signatures: module Functions where a :: Int -> [Int] a x = [x + 1, x + 2, x + 3] b :: Int -> [Int] b x = [x + 4, x + 5, x + 6] and in my main function I receive a command line arguments (a string) naming one of the functions in MyModule. I'd like to lookup and use of the functions by name: lookup :: String -> (Int -> [Int]) lookup "a" = MyModule.a lookup "b" = MyModule.b ... Is there a more idiomatic Haskell way of going about this? I can see there's a fundamental tension between a runtime choice of function name expressed as a string and the need for compile-time known function choice. Thanks, Stu

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change to wikibook?
by Ashley Smith 12 May '12

12 May '12

I was wondering if someone could explain this error? I typed in the sample code from the Haskell Beginning Wikibook to define a function for absolute value. I have GHC 7.0.4 for OSX. The error went away when I put parentheses around the negative value, so did the Haskell interpreter think I was entering - and a second argument, namely 10, when the parentheses weren't there? Thank you! *Main> let abs x = if x < 0 then -x else x *Main> abs 5 5 *Main> abs -10 And got the following error <interactive>:1:6: No instance for (Num (a0 -> a0)) arising from the literal `10' Possible fix: add an instance declaration for (Num (a0 -> a0)) In the second argument of `(-)', namely `10' In the expression: abs - 10 In an equation for `it': it = abs - 10

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