- Beginners - Haskell.org

possible incorrect indentation
by John Moore 26 Jan '10

26 Jan '10

Hi all, Will somebody please explain why this wont run, I sure the code is right? data Expression = Val Integer | Add Expression Expression | Subtract Expression Expression | Multiply Expression Expression | Divide Expression Expression deriving Show demo1 = (Add(Add(Add(Val 6)(Val 5))(Val 10))(Val 7)) evalStep :: Expression -> Expression evalStep (Val x)= (Val x) evalStep (Add x y) = case x of (Val a) -> case y of (Val b) -> Val (a+b) left -> Add x (evalStep y) right -> Add y(evalStep x) evaluate :: Expression -> Expression -- Base case evaluate (Val a) = Val a -- Recursive case evaluate e = do putStrLn "Evaluating one more step" e' <- return (evalStep e) putStrLn ("Result is "++(show e')) putStrLn "Do another step (y/n)? :" c <- getChar if (c=='y') then evaluate e' else putStrLn "OK, finished" Its suppose to print out the first step then ask you if you want the next step all the way to the end

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adding state handing to existing code
by Scott Thoman 25 Jan '10

25 Jan '10

Since I'm very new to Haskell I have what is probably a simple question yet I'm having trouble finding a clear example of how it works. The basic question is: how do I pass state through existing code without the intermediate code knowing about it. If I have, for example, several layers of function calls and the innermost function needs to access some state that is only "seeded" by the outermost function, how do I do that without the functions in between knowing about the additional state being threaded through them? I have a simple example (that may *not* be good idiomatic Haskell): -- process :: Integer -> Integer -> Integer process x y = 2 * x * y doit :: IO () doit = do printf "f x y = %d\n" $ process 42 43 main :: IO () main = do doit putStrLn "done" -- (I'm not totally sure about the type of "doit" but the code compiles and runs as expected) What I want to do is add some state handing to "process" to have it, say, count the number of times it's been called (putting threading/thread-local concerns aside for the moment). I'm trying to understand how to add state to "process" along the lines of: -- process :: Integer -> Integer -> State Integer Integer process x y = do s <- get put $ s + 1 return $ 2 * x * y -- but I want to only seed the state from "main" without "doit" having to change -- I can call "process" from "doit" like "(execState (process 42 43) 0)" but I want the initial state to be determined at the top level, from main. I have a feeling there's some kind of "ah ha" moment that I'm just not seeing yet. Any help or pointers to where I can look for myself would be greatly appreciated. Thanks in advance, -thor

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Critique my program
by David Frey 25 Jan '10

25 Jan '10

A couple of months ago, I was tutoring my cousin in an introductory computer science class. We talked about his assignment in which he had to implement a subset of the Battleship board game using Java. I decided it would be fun for me to try to complete the assignment in Haskell. I forgot about the project for a while, but I picked it up and finished it today. I have put the code on BitBucket (a Mercurial based code hosting site for those who don't know) at this location: http://bitbucket.org/dfrey/battleship/ I am hoping that some people from this list could review my code. I'm not very concerned with correctness. I'm more interested in how my code can be clarified through use of existing functions or standard Haskell idioms. If you don't have Mercurial installed or don't want to clone the repository, you can browse the source here: http://bitbucket.org/dfrey/battleship/src/ The original PDF from my cousin's assignment can be viewed here: http://bitbucket.org/dfrey/battleship/raw/81a01574f164/Assn4.pdf Feel free to submit your comments in any format. Patches, written comments, complete re-writes of the program, etc. are all welcome. Thanks, David Frey

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if statements
by John Moore 25 Jan '10

25 Jan '10

Hi I'm trying to get a if statement to work but not sure if this is the right approach, what I want to do in simple english is to evaluate one expression at a time for example. (3+2)+(4+5) I want it to return 5 + (4+5) then 5+9. data Expression = Val Integer | Add Expression Expression | Subtract Expression Expression | Multiply Expression Expression | Divide Expression Expression deriving Show evalStep :: Expression -> Expression evalStep (Val x)= (Val x) evalStep (Add x y) = do if x = isDigit then if y isDigit else evalStep x,y I thinking about using recursion down a tree left then right but that seems very complicated. Any thoughts welcome! John

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getChar and keyboard reading
by Matthias Güdemann 24 Jan '10

24 Jan '10

Hi, I wrote a program which should read the number of key presses in a given time. It forks a thread which calls getChar and increases a MVar each time getChar suceeds. After a delay (here 2s) the MVar is read, its value is printed and the program exits. import Control.Concurrent increaseEveryPush mVCount = do _ <- getChar inh <- takeMVar mVCount putMVar mVCount (inh + 1) increaseEveryPush mVCount main = do x <- newMVar 0 forkIO (increaseEveryPush x) threadDelay 2000000 inh <- takeMVar x putStr $ "\n" ++ (show inh) ++ " times button pushed\n" But this does not work, as getChar does not succeed if there is no EOF given (^D). It normally just exits with 0 as value in the MVar. Is there a way toread the keyboard without having to terminate with EOF? regards, Matthias

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Re: Re: testing and the culture of Haskell
by Adrian Adshead 23 Jan '10

23 Jan '10

OK, so I get the idea that all functions are pure because they just return actions. So as I understand it now... All functions are pure. The evaluation of all functions not declared as IO results in no side effects. The evaluation of all functions declared as IO may or may not have side effects. But how can the function greetAdrian from the example below not be an IO operation? From: Maciej Piechotka <uzytkownik2 <at> gmail.com> Subject: Re: Re: testing and the culture of Haskell Newsgroups: gmane.comp.lang.haskell.beginners Date: 2010-01-23 00:31:42 GMT (2 hours and 52 minutes ago) On Fri, 2010-01-22 at 20:59 +0000, Adrian Adshead wrote: > >All Haskell functions are pure without exception. For example: > > > >greet :: String -> IO () > >greet name = putStrLn $ "Hello, "++name > > > >This is a pure function from String to IO (). This function (like all > >Haskell functions) has no side effects. Its return value of type IO () > >merely _represents_ an IO action. The runtime system knows how to act > >on this representation. > > > >This also means that there is no such thing in Haskell as marking a > >function as side-effecting. > > > >This distinction may be subtle, but it's important. > > > > > >Steve > > Steve, > > Please could you clarify this for me since you are making exactly > the opposite assertion than I have understood. > > I am confused by you stating "All Haskell functions are pure > without > exception.". > > Pure functions have no impact on 'anything'. They take input > parameters (which they don't change) and return exactly the > same result whenever the same input parameters are given. > > >greet :: String -> IO () > >greet name = putStrLn $ "Hello, "++name > > This example you gave is not a pure function since it does have > the side effect that the screen is changed by outputting the string > "Hello, " and the name passed in. > > greatAdrian :: String greetAdrian = let x = greet "Adrian" in x `seq` f x greet can be consider a pure function and value IO () is evaluated by seq. IO () represents an action(s) not execution of action(s). If f of x does not use any tricks nothing will be printed. IO a value it can be: - cast unsafely into a. However I guess we omit this shame for a moment - binded with other action. But the resultant type is now again IO b. So we still get a something - returned as main. Then we might consider whole Haskell program as metalanguage which returns single thing - other program. In similar way as: type PythonProgram = String main :: PythonProgram main = "print \"Hello World\"" is pure add_impure is pure. What we do with the result is other thing. Regards

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different function implementations for different values of typevar - is it possible?
by Sergey Mironov 23 Jan '10

23 Jan '10

Hi all. I am writing a Monad intace for my own Writer-like data type "HtmlWriter" (see code below). The main HtmlWriter's difference is in return() function. I'd like it to convert user data (a) to html. But here come troubles. In general, there are two cases: 1) (HTML a) exists 2) (HTML a) doesn't exist So, i need to write different implementations of return() for both cases. Something like this: 1) return a = HtmlWriter (a, toHtml a) 2) return a = HtmlWriter (a, toHtml (show a)) I can't find a way to do it. Is it possible somehow? Please, help. Here is the code: ------------- import Control.Monad import Text.XHtml.Transitional newtype HtmlWriter a = HtmlWriter { runHtmlWriter :: (a, Html) } instance Monad (HtmlWriter) where return a = HtmlWriter (a, toHtml a) -- ??? return a = HtmlWriter (a, toHtml (show a)) m >>= k = HtmlWriter $ let (a, w) = runHtmlWriter m (b, w') = runHtmlWriter (k a) in (b, w `mappend` w') ------- Thanks, Sergey.

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Re: [Haskell-beginners] QuickCheck 2 entering infinite loop
by Dominic Mulligan 23 Jan '10

23 Jan '10

Dear Brent, Thanks for the reply. The type of args is [Type a]. Ascribing a specific type, like so: prop_InjectTypeConFreeVars :: [Type String] -> Bool prop_InjectTypeConFreeVars args = (freeTypeVars . (tyCon "Fun")) args == (concatMap freeTypeVars args) Leads to the same problem; quickCheck is brought to its knees after about 6 tests. Regards, Dominic

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Re: [Haskell-beginners] if statements
by Roel van Dijk 23 Jan '10

23 Jan '10

Hi John, The @ in a pattern is called an 'as-pattern'. They are explained in this tutorial: http://www.haskell.org/tutorial/patterns.html Very briefly they assign a name to a pattern. I'm not sure why you get a type error. I could load the code into GHCI without problems. When the compiler complains about missing instanced for Num than it usually means you did something like (x + y) where x and y are of type E without E having an instance for Num. For example (Val 3) + (Val 4) would result in such an error. On Sat, Jan 23, 2010 at 7:32 PM, John Moore <john.moore54(a)gmail.com> wrote: > Hi Roel, > This isn't working, what does @ stand for in > evalStep (Add x@(Val _) y) = Add x (evalStep y) > Maybe this is a mistype > It's asking me for a instance declaration for (num expression) > > John

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Stack Overflow with foldl'
by Cyril Pichard 23 Jan '10

23 Jan '10

Hi, I don't understand why the following code always result in a Stack space overflow error. I'm using foldl' and think the main algorithm is tail recursive. Am I wrong ? I've read the http://www.haskell.org/haskellwiki/Stack_overflow Any help would be greatly appreciated. Thanks module Main where import Data.List -- Data structures data Point = Point { x::Double, y::Double, z::Double } deriving(Eq, Show) data Triangle = Triangle { p1::Point, p2::Point, p3::Point } deriving(Eq, Show) data BBox = BBox { bbmax::Point, bbmin::Point } deriving(Eq,Show) -- Construct a default triangle newTriangle :: Triangle newTriangle = Triangle (Point 1 0 0) (Point 0 1 0) (Point 0 0 1) -- Construct a default BBox newBBox = BBox (Point 0 0 0) (Point 0 0 0) -- Defines min and max for Point instance Ord Point where max a b = Point ( max (x a ) (x b) ) ( max (y a) (y b) ) ( max (z a) (z b) ) min a b = Point ( min (x a ) (x b) ) ( min (y a) (y b) ) ( min (z a) (z b) ) -- Create a list of triangles makeTriangles :: Double -> [Triangle] makeTriangles 0 = [] makeTriangles n = [ Triangle (Point v 0 0) (Point 0 v 0) (Point 0 0 v) | v <- [1..n] ] -- Compute the bounding box of a triangle boundingBox :: Triangle -> BBox boundingBox t = BBox pmax pmin where pmax = max (p3 t) (max (p1 t) (p2 t)) pmin = min (p3 t) (min (p1 t) (p2 t)) -- Compute the bounding box of two bounding boxes bboxAddb :: BBox -> BBox -> BBox bboxAddb (BBox bb1max bb1min) (BBox bb2max bb2min) = BBox (max bb1max bb2max) (min bb1min bb2min) -- Compute the bounding box of a triangle and a bounding box bboxAddt :: BBox -> Triangle -> BBox bboxAddt b t = bboxAddb b (boundingBox t) main = do print $ foldl' bboxAddt newBBox triangles; where triangles = makeTriangles 610000 error returned : Stack space overflow: current size 8388608 bytes. Use `+RTS -Ksize' to increase it. ghc version 6.10.3 on MacOsX Snow Leopard

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