- Beginners - Haskell.org

folds -- help!
by 7stud 10 Mar '09

10 Mar '09

This is an example that shows how foldl and foldr work (from RWH p.93-94): foldl (+) 0 (1:2:3:[]) == foldl (+) (0 + 1) (2:3:[]) == foldl (+) ((0 + 1) + 2) (3:[]) == foldl (+) (((0 + 1) + 2) + 3) [] == (((0 + 1) + 2) + 3) foldr (+) 0 (1:2:3:[]) == 1 + foldr (+) 0 (2:3:[]) == 1 + (2 + foldr (+) 0 (3:[]) == 1 + (2 + (3 + foldr (+) 0 [])) == 1 + (2 + (3 + 0)) The book says on p.94: ----- The difference between foldl and foldr should be clear from looking at where the parentheses and the empty list elements show up. With foldl, the empty list element is on the left, and all the parentheses group to the left. With foldr, the zero value is on the right, and the parentheses group to the right. ---- Huh? With foldl, the only empty list element I see is on the right. Initially, it looked to me ike they did the same thing, and that the only difference was the way they called step. I think "step" is a horrible, non-descriptive name, so I'm going to use "accFunc" instead: foldl calls: accFunc acc x foldr calls: accFunc x acc So it looks like you can define a function using either one and get the same result. Here is a test: --I am going to use odd for pfunc and [1, 2, 3] for xs: myFilter1 pfunc xs = foldl accFunc [] xs where accFunc acc x | pfunc x = acc ++ [x] | otherwise = acc myFilter2 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = acc ++ [x] | otherwise = acc *Main> myFilter1 odd [1, 2, 3] [1,3] *Main> myFilter2 odd [1, 2, 3] [3,1] Hmmm. So there is a difference. foldr appears to grab elements from the end of the list. Therefore, to get the same result from the function that uses foldr, I did this: myFilter3 pfunc xs = foldr accFunc [] xs where accFunc x acc | pfunc x = x : acc | otherwise = acc *Main> myFilter3 odd [1, 2, 3] [1,3] But then RWH explains that you would never use foldl in practice because it thunks the result, which for large lists can overwhelm the maximum memory alloted for a thunk. But it appears to me the same thunk problem would occur with foldr. So why is foldr used in practice but not foldl?

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installing cabal with 6.10 - ubuntu
by B 10 Mar '09

10 Mar '09

So I was wanting to try out some code that relies on 6.10, I'm using ubuntu 8.04. Also, I'd like to have cabal installed so I can get and install packages easily. I found some directions online that said that compiling from source might be a good way to go. I downloaded and compiled ghc 6.10.1, installing it locally. The directions I used: ----------- directions ------------ I use 6.8.3 and 6.10.1 on ubuntu without problems. I always build ghc from source. It takes a bit of cpu time, but not much effort on my part. Assuming I've got an older version of GHC in my path, here's all I do: tar -xjf ~/Downloads/ghc-6.10.1-src.tar.bz2 tar -xjf ~/Downloads/ghc-6.10.1-src-extralibs.tar.bz2 cd ghc-6.10.1/ echo "HADDOCK_DOCS = YES" >mk/build.mk ./configure --prefix /usr/local/ghc-6.10.1 make -j sudo make install ------- end of directions -------- Ok so I did this. It works! Then, wanting to keep things simple, I removed ghc 6.8.3 so I'd only have one compiler and libraries around - lessening the chance of wierd conflicts. Done. Now to install cabal-install. Directions I got: - unpack cabal-install-0.6.2.tar.gz somewhere - cd into the resulting directory and run "sh bootstrap.sh" - add "~/.cabal/bin" to you path. Unfortunately, this doesn't work for me. bootstrap.sh requires parsec and network. parsec, in turn, requires mtl. - installed mtl - installed parsec Ok. Now when I go to build network I get this: Could not find module `Data.Generics': it is a member of package base-3.0.3.0, which is hidden According to google, the above message actually indicates that 'syb' is needed, the bug is here: http://hackage.haskell.org/trac/ghc/ticket/2980 - installed syb-0.1.0.0. - still get the same message. At this point I'm stuck! I should say that I'm using this to install all these things: runhaskell Setup configure --user runhaskell Setup build runhaskell Setup install So everything is installed 'locally' and not globally. Thanks for any help with this - I feel like this is harder than it is supposed to be, what am I missing??

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Top beginner mistakes
by 7stud 09 Mar '09

09 Mar '09

Here are my top three: 1) Putting an equals sign after a function name when using guards: myfunction x y = | x < 2 = "a" | y > 20 = "b" | otherwise = "c" dhask.hs:2:4: parse error on input `|' Failed, modules loaded: none. 2) Writing "if" in front of the guard conditions: myfunction x y = | if x < 2 = "a" | if y > 20 = "b" | otherwise = "c" dhask.hs:2:25: parse error on input `=' Failed, modules loaded: none. 3) Not putting parentheses around arguments of the form x:xs dosomething x:xs = head xs dhask.hs:1:0: Parse error in pattern Failed, modules loaded: none. In all three cases, the error messages don't help spot the problem.

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Log_b(a)
by Matthew J. Williams 08 Mar '09

08 Mar '09

Dear friends, What is the Haskell equivlent of log_b(a)? To clarify, log of 36 to the base 1.5. Sincerely Matthew J. Williams

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compiling first program --> warning
by 7stud 08 Mar '09

08 Mar '09

When compiling my first haskell program (the program on p. 71-72 of RWH), which I have in a file called hhask.hs, I got the following warning: $ ghc --make hhask [1 of 1] Compiling Main ( hhask.hs, hhask.o ) Linking hhask ... /usr/bin/ld: warning -F: directory name (/Users/me/Library/Frameworks) does not exist The program works fine, but I'm wondering about that warning. Does that mean I installed haskell in the wrong place? I installed haskell only last week on osx 10.4.11, and here is what I did: 1) I downloaded GMP.framework and GNUreadline.framework, which my mac automatically unzipped and placed on my desktop. I then dragged the resulting two folders into Library/Frameworks as per the instructions at: http://www.haskell.org/ghc/download_ghc_682.html#macosxintel 2) I downloaded ghc-6.8.2 3) I unzipped and untared into: /Users/me/my_tar_extractions 4) I cd'ed into the newly created ghc-6.8.2 folder. The full path to that folder is: /Users/me/my_tar_extractions/ghc-6.8.2 5) I read the INSTALL document in the ghc-6.8.2 folder. 6) I ran the command: $ ./configure 7) Then I rand the command: $ sudo make install 8) At the end of the install output, I got a message that said: ------------- Installation of ghc-6.8.2 was successful. To use, add /usr/local/bin to your PATH. Warning: this binary distribution does NOT contain documentation! -------------- 9) I appended /usr/local/bin onto the PATH in ~/.bash_profile.

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Using The Cabal that comes with GHC 6.10 on Windows
by Glenn 08 Mar '09

08 Mar '09

Hi, I'm sure this should be very simple, but I can't find any obvious documentation in the GHC doco that tells me how to get started with Cabal. I've installed GHC 6.10.1, which comes with Cabal. However, it doesn't seem to come in executable form ? So, I'm not sure what I need to do - build the executable, presumably ? I could download the prebuilt executable version of Cabal, but it seems pointless to include it in GHC, and then go download another one. Sorry for this stupid question ! Could anyone give me a hint on how to proceed ? Thanks ! :)

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defining own arbitrary :: Gen Int
by ben 07 Mar '09

07 Mar '09

Hello, I would like to define my own arbitrary which should only give naturals, i. e. which should behave like instance Arbitrary Integer where arbitrary = elements ([0..]). This definition gives a "multiple definition" error. Is there a way to hide the definition of arbitrary::Gen Int of QuickCheck, such that I can use my own definition ? Thanks a lot in advance, ben

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Basic experimenting with monads
by Patrick LeBoutillier 05 Mar '09

05 Mar '09

Hi all, I'm pretty much a haskell newbie (reading RWH at chapter 15) and I'm trying to understand monads better, more specifically the mechanics of >>= chains. For example, take the following code: skip :: IO a -> IO String skip a = return "?" main = do putStrLn "1" n <- skip $ do putStrLn "2" putStrLn "3" putStrLn $ "skipped " ++ n putStrLn "4" When I run it, it prints, as expected: 1 skipped ? 4 My understanding is that what is really "passed" to the skip function is something like this: putStrLn "2" >> putStrLn "3" Here is my question: is it possible for the skip function to return the number of actions that were skipped? In other words, is it possible to go through the chain above without executing the actions but counting them instead? It seems to me it should be possible to do this, in a way similar to what happens when executing the Maybe monad (a return value of Nothing triggers the return of other Nothing values, but the full length of the chain is still executed (RWH, p.38)). Am I on the right track here? Thanks a lot, Patrick -- ===================== Patrick LeBoutillier Rosemère, Québec, Canada

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let indenting problems
by 7stud 04 Mar '09

04 Mar '09

I get indenting errors with the following attempts to use let: ----- import Data.List (sortBy) mySort xs = sortBy myCompare xs let lx = length x ly = length y in where myCompare x y | lx < ly = LT | lx == ly = EQ | lx > ly = GT ------------------ import Data.List (sortBy) mySort xs = sortBy myCompare xs where myCompare x y let lx = length x ly = length y in | lx < ly = LT | lx == ly = EQ | lx > ly = GT ------------- import Data.List (sortBy) mySort xs = sortBy myCompare xs where myCompare x y let lx = length x ly = length y in | lx < ly = LT | lx == ly = EQ | lx > ly = GT ------------- This is what I ended up with: ------- import Data.List (sortBy) mySort xs = sortBy myCompare xs where myCompare x y | lx < ly = LT | lx == ly = EQ | lx > ly = GT where lx = length x ly = length y ------- But I would like to know how to use let.

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infix and bind pseudonym
by Michael Easter 04 Mar '09

04 Mar '09

Folks, I have a simple type called Beverage and an example that allows me to construct Maybe Beverage types. Here are some type signatures for example functions: request :: String -> Maybe Beverage addMalt :: Beverage -> Maybe Beverage I have defined a chain function like so: chain :: (Maybe a) -> (a -> Maybe b) -> (Maybe b) chain = (>>=) I can do this: (chain (request "beer") addMalt) and request "beer" `chain` addMalt I think I understand why, as I use the back-ticks for infix. However, I don't have to do that for the true bind function, (>>=) request "beer" >>= addMalt I would like to use chain in this way -- that is without back-ticks. I'm not sure how... Is there something I'm missing? thanks Mike ps. Thanks to everyone for the great discussion on IO (re: previous question) -- ---------------------- Michael Easter http://codetojoy.blogspot.com: Putting the thrill back in blog http://youtube.com/ocitv -> Fun people doing serious software engineering

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