March 2009 - Beginners

question from Yet Another Haskell Tutorial
by Michael Mossey 30 Mar '09

30 Mar '09

I have a question about a problem in Yet Another Haskell Tutorial (problem 7.1). My answers seems to disagree with Hal's, and it fact something looks wrong in Hal's answer (maybe it's an error in his paper). The problem is to write the following function in "point-free" style: func5 f list = foldr (\x y -> f(y,x)) 0 list Here's how I approached it. It's easy to drop 'list' by the eta-reduction (using partial application): func5 f = foldr (\x y -> f(y,x)) 0 But how to get rid of f? First I reasoned that f is a function of a tuple. That is, it is not curried. So to curry it: func5 f = foldr (\x y -> (curry f) y x) 0 The second argument to foldr is obviously flipping the arugments, so func5 f = foldr (flip $ curry f) 0 Now I want to use the eta-reduction again, but I have to transform this expression into something that takes f as its last argument instead of the 0. I can use flip again, this time on foldr: func5 f = flip foldr 0 (flip $ curry f) Now f can be dropped: THE FINAL ANSWER: func5 = flip foldr 0 . flip. curry This works, in a couple of my test cases. Now Hal gives this as the answer: func5 = foldr (uncurry $ flip f) 0 The first problem is that there's no argument f, though he refers to it. So maybe he meant func5 f = foldr (uncurry $ flip f) 0 But more problems. He's applying flip to f, but f takes only one argument (a 2-tuple). Then he's uncurry-ing it, but I thought it needed currying, not uncurry-ing. Can anyone figure out what Hal is up to, or does it look like a simple mistake? Thanks, Mike

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numerical types, the $ operator
by TG 30 Mar '09

30 Mar '09

Hi all I'm trying to understand the following simple function >-- | fractional part of number. >frac :: (RealFrac a) => a -> a >frac x = x - fromInteger . floor $ x which apparently is wrong. Whereas this is ok >frac x = x - fromInteger (floor x) Is the 1st one wrong because it is trying to apply the _whole_ 'left of $' to the 'x' on the right? How would an experienced guy write this without parentheses? Moreover, I've put the 'RealFrac' by looking at ":t floor". What kind of class constraint whould you put for doing eg: >frac x = x - fromInteger (floor (sqrt x) ) since 'floor' takes fractional and 'sqrt' takes RealFrac? Some kind of super-class? Thank you. PS Seems that types are half the language, if not more .. -- TG cowscanfly(a)airpost.net -- http://www.fastmail.fm - Access your email from home and the web

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Simple laplace solver
by Edward Z. Yang 29 Mar '09

29 Mar '09

Hello all, As my first foray into the wonderful world of Haskell programming, I wrote a simple program that solves Laplace equations. The program is 44 lines, and I would greatly appreciate comments on how I could make it better, with regards to Haskell idioms, performance and readability. Cheers, Edward import Data.Array data Node = Node { isBoundary :: Bool, value :: Double } instance Show Node where show (Node True value) = show value ++ "*" show (Node False value) = show value blankArray :: Ix i => (i, i) -> a -> Array i a blankArray (a, b) x = array (a, b) [(i, x) | i <- range (a, b)] fillBox :: Ix i => (i, i) -> a -> Array i a -> Array i a fillBox (a, b) value m = m // [(i, value) | i <- range(a, b)] average :: Fractional a => [a] -> a average list = sum list / fromIntegral (length list) solve :: Array (Integer, Integer) Node -> Array (Integer, Integer) Node solve m = if maximum deltas > 0.0001 then solve n else n where n = array (min, max) [(i, solveNode i) | i <- range(min, max)] deltas = map delta $ zip (values m) (values n) delta (x, y) = abs (x - y) values a = map value $ elems a (min, max) = bounds m solveNode (a, b) = if isBoundary $ node then node else averageNode where node = m ! (a, b) averageNode = Node False $ average [value $ m ! i | i <- [(a+1,b),(a-1,b),(a,b+1),(a,b-1)]] myArray = fillBox ((3,3), (5,5)) (Node True 100) $ fillBox ((1,1), (8,8)) (Node False 0) $ blankArray ((0,0), (9,9)) (Node True 0)

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Type inference question
by Zachary Turner 29 Mar '09

29 Mar '09

Given the following two function definitions: let func4 l = map (\y -> y+2) (filter (\z -> z `elem` [1..10]) (5:l)) let func4_pf = map (+2) . filter (`elem` [1..10]) . (5 :) which are equivalent, why does the first one have a type of func4 :: (Num a, Enum a) => [a] -> [a] while the second one has a type of func4_pf :: [Integer] -> [Integer] Shouldn't the types be the same?

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how do you use a package (.cabal)?
by 7stud 28 Mar '09

28 Mar '09

In chapter 5 in RWH, they show you how to create, configure, and install a package. This is the final step: ...mypretty$ runghc Setup install Installing: /usr/local/lib/mypretty-0.1/ghc-6.8.2 Setup: /usr/local/lib/mypretty-0.1: createDirectory: permission denied (Permission denied) .../mypretty$ sudo runghc Setup install Password: Installing: /usr/local/lib/mypretty-0.1/ghc-6.8.2 Registering mypretty-0.1... Reading package info from "dist/installed-pkg-config" ... done. Saving old package config file... done. Writing new package config file... done. $ ghc-pkg list /usr/local/lib/ghc-6.8.2/package.conf: Cabal-1.2.3.0, GLUT-2.1.1.1, HUnit-1.2.0.0, OpenAL-1.3.1.1, OpenGL-2.2.1.1, QuickCheck-1.1.0.0, array-0.1.0.0, base-3.0.1.0, bytestring-0.9.0.1, cgi-3001.1.5.1, containers-0.1.0.1, directory-1.0.0.0, fgl-5.4.1.1, filepath-1.1.0.0, (ghc-6.8.2), haskell-src-1.0.1.1, haskell98-1.0.1.0, hpc-0.5.0.0, html-1.0.1.1, mtl-1.1.0.0, mypretty-0.1, network-2.1.0.0, old-locale-1.0.0.0, old-time-1.0.0.0, packedstring-0.1.0.0, parallel-1.0.0.0, parsec-2.1.0.0, pretty-1.0.0.0, process-1.0.0.0, random-1.0.0.0, readline-1.0.1.0, regex-base-0.72.0.1, regex-compat-0.71.0.1, regex-posix-0.72.0.2, rts-1.0, stm-2.1.1.0, template-haskell-2.2.0.0, time-1.1.2.0, unix-2.3.0.0, xhtml-3000.0.2.1 The mypretty package is in that list. But what does that do for me? How do I access that package? The .cabal file has this text in it: library Exposed-Modules: SimpleJSON PrettyJSON Prettify Build-Depends: base >= 2.0 But I created a Main.hs file that imported the "exposed modules": module Main () where import Prettify import PrettyJSON import SimpleJSON main = let s = JString "hello world" aDoc = renderJValue s in printPrettified aDoc --JString is a JValue value constructor and JValue is defined in SimpleJSON.hs --renderJVAlue is a function defined in PrettyJSON.hs, and it returns a data type that is defined in Prettify.js --printPrettified is a function defined in PrettyJSON.hs This is what happens: $ ghc -c Main.hs ~/2testing/dir3$ ghc -o test Main.hs compilation IS NOT required /usr/libexec/gcc/i686-apple-darwin8/4.0.1/ld: warning -F: directory name (/Users/me/Library/Frameworks) does not exist /usr/libexec/gcc/i686-apple-darwin8/4.0.1/ld: Undefined symbols: ___stginit_myprettyzm0zi1_Prettify_ ___stginit_myprettyzm0zi1_PrettyJSON_ ___stginit_myprettyzm0zi1_SimpleJSON_ _myprettyzm0zi1_PrettyJSON_printPrettified_closure _myprettyzm0zi1_PrettyJSON_renderJValue_closure _myprettyzm0zi1_SimpleJSON_JString_static_info collect2: ld returned 1 exit status I googled cabal tutorials, but I can't find anything that tells me how to access an installed package.

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Why doesn't ghci load this?
by Jeff Lasslett 28 Mar '09

28 Mar '09

Hi, I've been making my way through the exercises in the book "Real World Haskell". Below is my solution to an exercise asking for an implementation of groupBy in term sof a foldr. I haven't tested it extensively but is seems to work. GHCI rejects it though, if I uncomment the type signature for the 'step' function. Could someone please help me to understand why that is? Thanks, Jeff myGroupBy :: (a -> a -> Bool) -> [a] -> [[a]] myGroupBy p xs = foldr step [[]] xs where -- step :: a -> [ [ a ] ] -> [ [ a ] ] step x acc@( ys : yss ) | null ys = [ x ] : [] | p x ( head ys ) = ( x : ys ) : yss | ( p x ( head ys ) ) == False = [ x ] : acc

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Tutorial/Book with Exercises
by Zachary Turner 28 Mar '09

28 Mar '09

Which book or tutorial has the best exercises? Whenever I learn a language I like to do tons of exercises, and in fact I keep a list of books with great exercises. I currently have Real World Haskell and Programming in Haskell (Hutton), but I feel the exercises in these books are lacking. Examples of books with great exercises are ANSI Common Lisp (Graham), Developing Applications with Ocaml (O'Reilly), The Algorithm Design Manual (Skiena), and and Enumerative Combinatorics by Stanley (this book is the king of all books when it comes to exercises). They shouldn't be impossible, but at the same time exercises like "what are the types of the following expressions" are pretty silly. I think a good book should be broken into chapters where it takes at least 10 to 20 times as long to complete the exercises at the end of each chapter than it did to read the chapter and understand the material in it.

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Getting source libraries
by Zachary Turner 27 Mar '09

27 Mar '09

Suppose I installed a GHC binary package, and now I decide I want source for all my libraries. What is the easiest way to go about doing this? In theory I could download every source package individually and unpack the source into my directory tree, but this is really a last resort since there are a lot of libraries.

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pattern matching to data inside a list
by Michael Mossey 27 Mar '09

27 Mar '09

Is there a way to pattern match to a list argument and get a list? For example, I have this: -- A LayoutItem has a center position (point), and the four ints are -- "left width", "rigth width", "top height", and "bottom height" data LayoutItem = LayoutItem Point Int Int Int Int deriving Show -- This computes the left width of a group of composited LayoutItem. compositeLeftWidth :: [LayoutItem] -> Int compositeLeftWidth items = let itemsPosX = [ x | LayoutItem (Point (x,_)) _ _ _ _ <- items ] itemsLW = [ lw | LayoutItem _ lw _ _ _ <- items ] z = zipWith (\x y -> x-y+1) itemsPosX itemsLW in (minimum z) - 1 What I'm wondering is if I could somehow do something like compositeLeftWidth [ LayoutItem Point( x, _ ) lw _ _ _ ] = ... and that x and lw would each be the lists I'm calling itemsPosX and itemsLW above. Thanks, Mike PS Let me repeat my request to find out how to install SOE "on the library path." I know I'm impatient about this, but it seems simple enough, and it's driving me crazy that I can only write software in the same directory where I've installed SOE.

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question about styles of recursion
by Michael Mossey 27 Mar '09

27 Mar '09

In RWH, in the exercises at the end of the book, I was told to write a function that averages the integer values in a list. I wanted to do this using on the tools we had been presented, which did not include 'length'. So I thought of writing a recursive function in which each case passes an accumulator of the "sum so far" as well as a count of node "up to the point", and the base case does the actual division. I was wondering if there is a better way of doing it (using the just ideas up to chapter 3, so no length, no higher order functions, no foldr/foldl or map). myAvg' :: Int -> [Int] -> [ Double ] -> Double myAvg' sum count [] = sum / fromIntegral count myAvg' sum count (x:xs) = myAvg' (x + sum) (n + 1) xs myAvg :: [Double] -> Double myAvg xs = myAvg' 0 0 xs Thanks, Mike

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