August 2014 - Beginners

understanding curried function calls
by Dimitri DeFigueiredo 21 Aug '14

21 Aug '14

Here's a simple exercise from Stephanie Weirich's class [1] that I am having a hard time with. consider doTwice :: (a -> a) -> a -> a doTwice f x = f (f x) what does this do? ex1 :: (a -> a) -> a -> a ex1 = doTwice doTwice At least, it is clear that there is a parameter to doTwice missing. So, I wanted to do: ex1 y = (doTwice doTwice) y but this gets me nowhere as I don't know how to apply the definition of doTwice inside the parenthesis without naming the arguments. What is the systematic way to evaluate these expressions? I actually got really stumped when I considered. ex2 :: (a -> a) -> a -> a ex2 = doTwice doTwice doTwice doTwice I assume this is not the same as ex2 = (doTwice doTwice doTwice) doTwice what's being applied to what here!? Are there any resources with many practice exercises like this one? Thanks, Dimitri [1] http://www.seas.upenn.edu/~cis552/lectures/Lec2.html

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Detecting duplicate code by AST?
by Frank 21 Aug '14

21 Aug '14

Hi, Can Anyone point Me in the direction of a tool/method of detecting code duplicates based on analysis of abstract syntax trees (ASTs)? If not, is there a way to dump the ASTs of a Haskell project? Is this something for Template Haskell to handle? Sincerely, Frank D. Martinez -- P.S.: I prefer to be reached on BitMessage at BM-2D8txNiU7b84d2tgqvJQdgBog6A69oDAx6

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Question on evaluating function compostion
by Chul-Woong Yang 21 Aug '14

21 Aug '14

Hi, all I'm having trouble in understanding function evaluation in Haskell. I came across the following line, which is somewhat cryptic to me. (liftM . (+)) 1 [2] Could you explain how the expression evaluates? I thought that to evalutate two composed functions, I should apply right function to get a result and then apply left function with the result. e.g. f.g x y = f (g x y) = f z = result So I guessed that Haskell evaluated above expression as follows: (liftM . (+)) 1 [2] ---> ((liftM . (+)) 1) [2] ---> (A) (liftM (+1)) [2] --> [3] Why did Haskell, however, not try to fully evaluate addition, like following? (liftM . (+)) 1 [2] ---> liftM ((+) 1 [2]) ---> error Does (f . g) x y z equal ((((f . g) x) y) z) in haskell? Any guide will be appreciated. Chul-Woong

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Disappointing thought on Haskell -- a simple space leak on "insert" is hard bug to catch
by Dimitri DeFigueiredo 20 Aug '14

20 Aug '14

Here's a problem I just found with Haskell (and maybe functional programming) that I would like to know how to avoid. Please don't get me wrong. I'm in *LUUV* with Haskell! :-) Exercise 2.3 of Chris Okasaki's book "Purely Functional Data Structures" shows that something as simple as an insert function in a binary tree can cause a space leak in Haskell. (and other functional languages as well) Here's a simple tree type: data Tree a = Empty | MkTree (Tree a) a (Tree a) Here's the version of insert with the space leak: -- ** SPACE LEAK! ** version -- this version unnecessarily copies the path to the inserted element even if -- the same element is already present in the tree. -- a million insertions of the same element will cause a million trees -- to exist whereone would suffice. treeInsert :: Ord a => a -> Tree a -> Tree a treeInsert e Empty = MkTree Empty e Empty treeInsert e t@(MkTree a x b) | e > x = MkTree a x (treeInsert e b) | e < x = MkTree (treeInsert e a) x b | otherwise = t Here's a version without the space leak: treeInsert :: Ord a => a -> Tree a -> Tree a treeInsert e t = case (treeInsert' e t) of Nothing -> t Just x -> x where treeInsert' :: Ord a => a -> Tree a -> Maybe (Tree a) treeInsert' e Empty = return $ MkTree Empty e Empty treeInsert' e (MkTree a x b) | e > x = do {r <- treeInsert' e b; return $ MkTree a x r} | e < x = do {r <- treeInsert' e a; return $ MkTree r x b} | otherwise = Nothing This is so tricky that I think it is worth questioning whether Haskell is helping here. First, just spotting that the first version leads to a space leak is not trivial. Second, the fix is so much uglier than the original version. It is disappointing to have to write it! Questions: 1) ** Is there a warning sign that would easily flag this space leak? ** 2) Is there a simpler fix? Cheers, Dimitri

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Using Traversal as a kind of pointer
by Elise Huard 19 Aug '14

19 Aug '14

Hi, I'm programming a small game - amongst my types I have a 'world' structure data World = World Background [Lifeform] I'm calculating collisions between my player and the lifeforms, and I'll need this result both to allow some actions and to modify the world as a result. I'm using FRP with elerea, so everything is defined as behaviours (signals), and the collision signal is used in the signal accounting of life spent and in the world signal. I was thinking of returning a Traversal on the lifeforms that are colliding so that I can both access the lifeforms concerned (when performing actions and life accounting) and modify them as a result (for the world). Does that sound reasonable as a use of lenses? Overengineered? Better solutions? Thank you, Elise Huard

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Data.Stream interleave implementation question
by Bryan Brady 18 Aug '14

18 Aug '14

I'm in the process of learning Haskell and am implementing my own Stream type. ( I'm working my way through homework problems I found at http://www.seas.upenn.edu/~cis194/hw/06-laziness.pdf ) One of the questions is: "Deﬁne the stream ruler :: Stream Integer which corresponds to the ruler function 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, . . . where the nth element in the stream (assuming the ﬁrst element corresponds to n = 1) is the largest power of 2 which evenly divides n. " I got pretty close, but ended up looking at Data.Stream to nail down my mistake. My original solution was: ----- data Stream a = Cons a (Stream a) deriving (Eq, Ord) streamRepeat :: a -> Stream a streamRepeat a = Cons a (streamRepeat a) interleaveStreams :: Stream a -> Stream a -> Stream a interleaveStreams (Cons a as) (Cons b bs) = Cons a (Cons b (interleaveStreams as bs)) ruler :: Stream Integer ruler = foldr1 interleaveStreams (map streamRepeat [0..]) ----- Attempting to evaluate 'ruler' in ghci does not finish. I knew something was wrong with my implementation of interleaveStreams because it worked when I expanded it out manually for a few steps. The way Data.Stream implements interleave is: interleaveStreams (Cons a as) bs = Cons a (interleaveStreams bs as) I don't understand why this works and why my original implementation doesn't. I figure if the latter works, the former should as well. Here is my reasoning: In the latter definition, Cons a (interleaveStreams bs as), (interleaveStreams bs as) is a thunk. The thunk should only be evaluated when it is needed. In my original definition, (interleaveStreams as bs) is a thunk. The difference is an extra Cons (e.g., Cons b). It seems like the additional Cons is causing problems, I just don't understand why. Can anyone point out what I'm missing? thanks, bb

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"Learn You a Haskell" question
by Frank 18 Aug '14

18 Aug '14

Hi, I'm reading "Learn You a Haskell..." and have a question about the chapter "Making Our Own Types and Typeclasses". On the 'Functor'/'Either' example, I feel completely lost. I don't see why the 'Left x' portion of 'Functor (Either a)' is simply 'Left x' and the document is not exactly clear. Any clarification would be most appreciated. Sincerely, Frank D. Martinez -- P.S.: I prefer to be reached on BitMessage at BM-2D8txNiU7b84d2tgqvJQdgBog6A69oDAx6

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How to show a record
by martin 17 Aug '14

17 Aug '14

Hello all, I have a record type like this data Payload = Pl { pType :: Char, accepts :: String, produces :: String, workarea :: String} deriving (Eq) an I want to write my own show function. I ended up with somethink like this show pl = concatMap ( $ pl) [show . pType, show . accepts, show . produces, show . workarea] That repetition of "show ." annoys me, but of course without it the elements of that list would have different types. Is there a way around it and how does haskell itself manage to show records?

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Re: [Haskell-beginners] Sandbox Confusion
by Britt Anderson 15 Aug '14

15 Aug '14

This works if one removes the -threaded option e.g. cabal exec ghc site.hs Is there a way to make cabal exec accept ghc options? --Britt > Message-ID: > <CAKFCL4V0cntKL6dvXe0Yh2cGf6JwEeHauQk2HEv3Hctz26JNpw(a)mail.gmail.com> > Content-Type: text/plain; charset="utf-8" > > On Thu, Aug 14, 2014 at 8:13 PM, Britt Anderson <britt(a)uwaterloo.ca> wrote: > > > cabal exec hakyll-init my-site > > and this works, and I > > cd ./my-site > > but the next step specified in the tutorial I can't figure out how to work. > > It is: > > ghc --make -threaded site.hs > > > > It fails because it cannot find the Hakyll module to import. > > > > I am hoping to learn what the proper command would be in a sandbox > > environment, and > > > You already used it once: put `cabal exec` in front. (Also, you no longer > need `--make` in current versions of ghc.) > > -- > brandon s allbery kf8nh sine nomine associates > allbery.b(a)gmail.com ballbery(a)sinenomine.net > unix, openafs, kerberos, infrastructure, xmonad http://sinenomine.net >

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Sandbox Confusion
by Britt Anderson 14 Aug '14

14 Aug '14

The cabal sandbox is confusing me. Specifically, trying to install hakyll in a sandbox. I begin with: cabal sandbox init cabal install -j hakyll and all goes well, then to run the shell script as specified in the tutorial here (http://jaspervdj.be/hakyll/tutorials/01-installation.html) cabal exec hakyll-init my-site and this works, and I cd ./my-site but the next step specified in the tutorial I can't figure out how to work. It is: ghc --make -threaded site.hs It fails because it cannot find the Hakyll module to import. I am hoping to learn what the proper command would be in a sandbox environment, and hopefully that will help me understand this sanboxing a bit better. It seems that haskell is developing faster than my beginners knowledge, and I am losing ground. I used to at least be able to install things.

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