- Beginners - Haskell.org

curry in a hurry
by prad 04 Jul '10

04 Jul '10

today i'm trying to understand curry and uncurry (in 3 parts with specific questions labelled as subsections of the parts). part 1 if we have f (x,y) = x + y this is a function working on a single argument (x,y) albeit composed of 2 parameters. however, since all functions are curried in haskell (read that here: http://www.haskell.org/haskellwiki/Currying) what is really happening is (f x) y or specific to this case (+ x) y since f ends up being (+) as defined. now if we write g = curry f we are naming a function whose purpose is to (f x) whatever comes its way allowing us to do neat things like: map (g 3) [4,5,6] [7,8,9] which we can't do by map (f 3) [4,5,6] because f :: (Num t) => (t, t) -> t meaning f - is a function ... the (Num t) - applies itself ... the => - to a Pair of (Num t)s - gives back a Num t 1.1 am i reading the type statement correctly? while g :: Integer -> Integer -> Integer means g takes an Integer, applies itself and that Integer to another Integer and computes another Integer. 1.2 how come these are Integer suddenly and not Num t? this is a problem because while i can do f (2.3,9.3) i get an error when i try (g 2.3) 9.3 ghci wants an instance declaration for (Fractional Integer) which puzzles me because g came about through currying f which is fine with fractions. part 2 now the next discovery is really strange to me. if i name f x y = x + y we see f :: (Num a) => a -> a -> a which looks like the curried form of f (x,y) in fact that's what it exactly is and i can do map (f 3) [4,5,6] [7,8,9] just as i did with g before!! which is what the wiki statement says too: f :: a -> b -> c is the curried form of g :: (a, b) -> c however, it starts by stating: "Currying is the process of transforming a function that takes multiple arguments into a function that takes just a single argument and returns another function if any arguments are still needed." http://www.haskell.org/haskellwiki/Currying 2.1 so does all this mean that f (x,y) is the function that takes multiple arguments and not a single argument as i initially thought and f x y is the function that actually takes a single argument twice? part 3 some of the above seems to be confirmed by looking at these types c x y = x + y c :: (Num a) => a -> a -> a so that's curried u (x,y) = x + y u :: (Num t) => (t,t) -> t so that's uncurried :t uncurry c uncurry c :: (Num a) => (a, a) -> a :t curry u curry u :: (Num a) => a -> a -> a but :t uncurry u uncurry u :: (Num (b -> c)) => ((b -> c, b -> c), b) -> c we're trying to uncurry something that is already uncurried and :t curry c curry c :: (Num (a, b)) => a -> b -> (a, b) -> (a, b) we're trying to curry something that is already curried 3.1 just what are these strange looking things and how should their types be interpreted? -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's

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'cabal install hdirect' can't find hdirect
by Larry Evans 04 Jul '10

04 Jul '10

hdirect is here: http://hackage.haskell.org/package/hdirect however, as shown by the following terminal session, cabal can't find it. What should I do to insteall hdirect? ~/download/haskell/libs $ cabal -V cabal-install version 0.6.0 using version 1.6.0.1 of the Cabal library ~/download/haskell/libs $ uname -a Linux evansl-desktop 2.6.24-24-generic #1 SMP Fri Sep 18 16:16:18 UTC 2009 x86_64 GNU/Linux ~/download/haskell/libs $ cabal install happy --dry-run Resolving dependencies... In order, the following would be installed (use -v for more details): mtl-1.1.0.2 happy-1.18.1 ~/download/haskell/libs $ cabal install hdirect --dry-run cabal: There is no package named hdirect ~/download/haskell/libs $

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Emacs tags
by Michael Mossey 04 Jul '10

04 Jul '10

(I asked this question on the logical email list to ask, haskell-mode, but got no reply, so I'm hoping someone here might know the answer.) I tried to use tags with emacs. I ran :etags in ghci, then tried to use C-c M-. inside Emacs to find the definition of a symbol. It reported "no source information available". I did some basic checking---a TAGS file does exist in the same directory as the file I was visiting, and contains some kind of entry for the specific symbol (whether the right syntax I don't know). I'm using haskell-mode 2.4 and emacs 21.3.1 on Windows XP. Also ghci 6.12.1. Thanks, Mike

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Where is hugs (or other) version of gofer modular-interpreter code?
by Larry Evans 03 Jul '10

03 Jul '10

Page 2 of: http://www.cs.yale.edu/~liang-sheng/popl95.ps.gz says the source code is at: nebula.cs.yale.edu/pub/yale-fp/modular-interpreter However, that's no longer present. Since Hugs is the successor to gofer, I was wondering if anyone could tell me how to get a copy of the corresponding hugs code. TIA. -Larry

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multreplace
by prad 02 Jul '10

02 Jul '10

so after being inspired by the continual excellent help here, i am trying my hand at writing a multiple replace using Useful.replace Useful.replace origStr pattStr replStr gives for Useful.replace "This original string" "orig" "very orig" "This very original string" what i want to do is something like multRepl origStr [patt1,patt2] [repl1,repl2] so we get from multRepl "This original string" ["orig","ing"] ["very orig","ucture"] "This very original structure" what i have come up with is: multRepl :: String -> [String] -> [String] -> [String] multRepl str ss rs = [U.replace str a b | (a,b) <- (zip ss rs)] this produces in 2 passes: ["This very original string","This very original structure"] to get the final result, i could just last multRepl but i would like to know if this is a good way to do things. i suppose going through the incremental steps is obligatory, but after printing out the result and seeing these steps, it feels like i'm producing throw away stuff - though perhaps that's just an illusion stemming from a similar function php5 has where you get the result without seeing anything behind the scenes. -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's

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upgrade Hackage show to QuickCheck 2 for lambdabot
by Mark Wright 02 Jul '10

02 Jul '10

Hi, I'm trying to upgrade Hackage show to QuickCheck 2, after applying the diffs below (which may not be correct, since I am a beginner), I am left which this error message: runghc ./Setup.hs build Preprocessing library show-0.3.4... Building show-0.3.4... [4 of 4] Compiling ShowQ ( ShowQ.hs, dist/build/ShowQ.o ) ShowQ.hs:104:20: Not in scope: `generate' Compilation exited abnormally with code 1 at Fri Jul 2 23:07:17 The error occurs in this method: tests :: Gen Result -> StdGen -> Int -> Int -> [[String]] -> IO String tests gen rnd0 ntest nfail stamps | ntest == 500 = done "OK, passed" ntest stamps | nfail == 1000 = done "Arguments exhausted after" ntest stamps | otherwise = case ok result of Nothing -> tests gen rnd1 ntest (nfail+1) stamps Just True -> tests gen rnd1 (ntest+1) nfail (stamp result:stamps) Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ++ reason result where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 The QuickCheck 1 generate method is near the bottom this page: http://hackage.haskell.org/packages/archive/QuickCheck/1.2.0.0/doc/html/Tes… but I can not find generate in QuickCheck 2. I am wondering if you have any ideas on how to fix it? I'm trying to package lambdabot on Solaris. I have already packaged the Haskell Platform and about 90 packages, they are in: http://pkgbuild.sourceforge.net/spec-files-extra/ Thanks very much, Mark here are the diffs: goanna% diff -wc show-0.3.4-orig/ShowQ.hs show-0.3.4/ShowQ.hs *** show-0.3.4-orig/ShowQ.hs Wed Jan 20 11:24:11 2010 --- show-0.3.4/ShowQ.hs Fri Jul 2 23:07:13 2010 *************** *** 12,22 **** --- 12,25 ---- import qualified Test.SmallCheck (smallCheck, Testable) import Test.QuickCheck + import Test.QuickCheck.Arbitrary import Data.Char import Data.List import Data.Word import Data.Int import System.Random + import Control.Exception (evaluate) + import Test.QuickCheck.Property (ok, stamp) type T = [Int] type I = Int *************** *** 23,36 **** --- 26,45 ---- instance Arbitrary Char where arbitrary = choose (minBound, maxBound) + + instance CoArbitrary Char where coarbitrary c = variant (ord c `rem` 4) instance Arbitrary Word8 where arbitrary = choose (minBound, maxBound) + + instance CoArbitrary Word8 where coarbitrary c = variant (fromIntegral ((fromIntegral c) `rem` 4)) instance Arbitrary Ordering where arbitrary = elements [LT,EQ,GT] + + instance CoArbitrary Ordering where coarbitrary LT = variant 1 coarbitrary EQ = variant 2 coarbitrary GT = variant 0 *************** *** 37,42 **** --- 46,53 ---- instance Arbitrary Int64 where arbitrary = sized $ \n -> choose (-fromIntegral n,fromIntegral n) + + instance CoArbitrary Int64 where coarbitrary n = variant (fromIntegral (if n >= 0 then 2*n else 2*(-n) + 1)) instance (Integral a, Arbitrary a) => Arbitrary (Ratio a) where *************** *** 48,53 **** --- 59,65 ---- else (b % a) else (a % b) + instance (Integral a, CoArbitrary a) => CoArbitrary (Ratio a) where coarbitrary m = variant (fromIntegral $ if n >= 0 then 2*n else 2*(-n) + 1) where n = numerator m *************** *** 87,93 **** Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ! ++ unlines (arguments result) where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 --- 99,105 ---- Just False -> return $ "Falsifiable, after " ++ show ntest ++ " tests:\n" ! ++ reason result where result = generate (((+ 3) . (`div` 2)) ntest) rnd2 gen (rnd1,rnd2) = split rnd0 goanna%

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the role of assignments
by prad 02 Jul '10

02 Jul '10

i'm trying to sort the assignment concept out in my mind. as i understand it so far, <- is for IO so you can do something like someIOVar <- readFile filename this will be of type IO String which is different from String as in let someStrVar = "this is a string" to work with an IO String you need to convert it into a String which seems to automatically happen inside a do structure as in: main = do tmplX <- readFile "zztmpl.htm" navbx <- readFile "zznavb.htm" let page = U.replace tmplX "NAVX" navbx are there some other ways to make IO String into String? also, it seems that assignment is different for the '=' in a program vs ghci for functions: sum x y = x + y (program) vs let sum x y = x + y (ghci) but not for strings and other things because you must always preface assignment of values with 'let': let a = 4 i suppose the let is there for variable assignments because such things necessitate a change of state and i've read that this is not desirable in functional programming - you have to work a little harder to do assignment than languages which just allow a = 4 b = 5 c = 6 etc in haskell, is it preferable to do something like this: var <- readFile fn someFunction var or someFunction (readFile fn) -- In friendship, prad ... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's

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functor declaration
by Michael Mossey 01 Jul '10

01 Jul '10

in the declaration of Functor: class Functor f where fmap :: (a -> b) -> f a -> f b does the compiler infer that f is a type constructor (not a type) because of the appearance of "f a" and "f b"? What I'm thinking is that some classes are classes of types, not type constructors. Like class CanMakeInt a where makeInt :: a -> Int In this case a is a type, not a type constructor. But there is no difference in the form of the first line of the class declaration. Thanks, Mike

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a bunch o' questions
by Michael Mossey 01 Jul '10

01 Jul '10

I have some questions about terminology re types and type constructors. If we have data Tree a = Branch (Tree a) (Tree a) | Leaf a Then is 'Tree' a type constructor? Is it correct to say: - 'Tree' is parameterized by one type, a - 'Tree a' is a type Regarding (->): - Is (->) a type constructor parameterized by two types? - Is (->) an operator? If so, it is the only operator that is also a type constructor? Thanks, Mike

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Re: [Haskell-beginners] a bunch o' questions
by Markus Läll 01 Jul '10

01 Jul '10

Leaf and Branch are type-constructors, and "Tree a" is a parametrized type. Constructor is what you actually use to construct trees, but the type (Tree a) you use to annotate values. Like in > let t = Branch (Leaf 1) (Leaf 2) :: Tree Int you constuct a tree with two leaves, and after two colons say, that the type of the tree is Tree Int Markus

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