- Beginners - Haskell.org

Platform for Mac OS 10.6
by Henry Lockyer 20 Apr '14

20 Apr '14

Hello people - trying to get back into Haskell again after previous flirtation which was long enough ago to forget almost everything, it seems. I appear to be lost in a mirror maze in trying to work out again how to bring my platform more up to date on Mac OS 10.6.8 (Snow Leopard). The macos link for download at http://www.haskell.org/platform/mac.html however does not seem to work. I do not get a working installer package just an "Unknown" 204mb file. ? Sniffing around via GHC pages I find this page: http://www.haskell.org/ghc/distribution_packages where it says "For Mac OS X we provide a native system installer, available on the download page for each release. We recommend using these installers " However the 'download page' link just takes you circularly back to http://www.haskell.org/ghc/download ? This page http://www.haskell.org/platform/index.html says the the current platform version is 2013.2.0 From my own past notes I should currently have platform 2011.4.0.0 I do definitely have GHC 7.0.4-x86_64 I also ran 'cabal update' as I dimly remembered that I should do this, and it indicated there was another version of cabal install available by doing a "cabal install cabal-install" but this complains: "$ cabal install cabal-install Resolving dependencies... cabal: cannot configure process-1.2.0.0. It requires base >=4.4 && <4.8 For the dependency on base >=4.4 && <4.8 there are these packages: . . ." I have base 4.3.1.0 (as indicated by GHC-PKG LIST). I thought refreshing the overall package level would be the most straightforward way to update to a consistent state as at least the first step. Hopefully I'm just having a bad day... I have two main motivations here (1) general refresh (2) get latest Open-GL stuff - in case that affects the answer. Can anyone recommend/remind/help with how best to approach this? Many thanks/ Henry

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failed to install HTTP-4000.2.4 when executing `cabal install cabal-install`
by hengchu Zhang 20 Apr '14

20 Apr '14

Hi Friends, When I tried to update cabal-install today by doing `cabal install cabal-install`, I got the following message due to failure in compiling Network/HTTP/Auth.hs, here is a screenshot of the process I am using Mac OS 10.8.4, and I installed haskell from homebrew. Does this mean something might have went wrong when I installed haskell? I have Xcode 5.1.1 and the latest command line tools installed. Does anyone know what might be going wrong? Thank you so much for your help! Best, Hengchu

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Learn You a Haskell! I have a few questions!
by Gilberto Melfe 19 Apr '14

19 Apr '14

Hi there to you all! I've been reading through the first chapter of "Learn You a Haskell" and I'd like to ask the community a few questions. Any help would be appreciated... -- ---------- ---------- ---------- ---------- ---------- All the standard Prelude functions that throw out an error when fed the []! head maximum ... Are there safe versions anywhere, or do we have to define them ourselves? -- ---------- ---------- ---------- ---------- ---------- This one is really important! I understand that for the definition of product to work as it is: product [] must be equal to 1 But what if we want to add a product to something else??? Shouldn't the result be Nothing? (I guess we would have to guard against this! But we must guard against the parameter of product being the empty list, anyway. Otherwise we risk adding 1 when there is nothing do multiply) (The same question arises with the functions and and or, and their boolean results, I think! Right?) -- ---------- ---------- ---------- ---------- ---------- -- ---------- Start Quote Names can't be enumerated. What comes after "John"? I don't know. -- ---------- End Quote "a" to "z" then "aa" to "zz" then "aaa" to "zzz" and so on! Is it to difficult or impossible to create a function that enumerates all possible strings? -- ---------- ---------- ---------- ---------- ---------- -- ---------- Start Quote To make a list with all the numbers from 20 to 1, you can't just do [20..1], you have to do [20,19..1]. -- ---------- End Quote Why is this? If the first was greater than the second it would just subtract! Right? -- ---------- ---------- ---------- ---------- ---------- -- ---------- Start Quote Watch out when using floating point numbers in ranges! Because they are not completely precise (by definition), their use in ranges can yield some pretty funky results. ghci> [0.1, 0.3 .. 1] [0.1,0.3,0.5,0.7,0.8999999999999999,1.0999999999999999] -- ---------- End Quote Can anyone explain me why it works for the first few values, and not "completely"? -- ---------- ---------- ---------- ---------- ---------- Any thoughts? Thank You Very Much in advance! Gilberto

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Learning Haskell: Web Project
by Ari King 17 Apr '14

17 Apr '14

Hi, I'm just starting out with Haskell and I thought a good way to learn/practice would be to re-implement a small web app. The web app I plan on re-implementing is made up of a restful (java) api with a angularjs client. The api and client is served by jetty. I'd appreciate suggestions on frameworks and/or tools that make the most sense as I start out. I was thinking of using scotty + warp + persistence for the api/serving of static content. As for tooling, I was thinking of vim with ghc-mod & syntastic. Thanks. -Ari

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recursive 'let' ?
by John M. Dlugosz 15 Apr '14

15 Apr '14

I understand that the definitions introduced by 'let' can be recursive, even mutually recursive among several names. Why would you want to do that? I saw contrived examples, and wonder why the authors never show a realistic example. let b = f a c c = f a b in ... I see that makes sense in light of lazy evaluation: b is really an alias for a (recursive) function, not a value that needs to find fixed points. Is this used for common idioms and problem-solving approaches in Haskell? --John

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Why getArgs is in the IO monad
by Mike Meyer 15 Apr '14

15 Apr '14

> > > From: "John M. Dlugosz" <ngnr63q02(a)sneakemail.com> > (And that begs the question of why getArgs needs to be monadic in the > first place. It > doesn't change its value; it's a strict constant at run-time, and not > knowing it at > compile time is my problem how?) > Actually, the value of the arguments can be changed - at least on some platforms. They are writable from C, if nothing else. What should getArgs do if some ffi changes the arguments before it's called? Mike

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general structuring of "foreach" logic in IO
by John M. Dlugosz 15 Apr '14

15 Apr '14

This works: main = do myargs <- getArgs mapM_ (\s -> putStrLn s ) myargs and imagine that the "body" will be substantial rather than just a putStrLn. My gut instinct is that the code ought to be arranged as: <any needed keywords or punctuation> and <the collection of items> <body to perform for every element ... ... > Meanwhile, there is no need to name the result of getArgs into myargs. So, getArgs is of type IO [String], and I want to apply that in the manner of a list. Without the Monad wrappers, plain map ( blah ) strings could be ( blah ) <$> strings, and in this particular case I don't see a reversed-arg version, although there is one for <*> (as <**>). But, for monad stuff in general there are reversed arrows for (most?) everything, and that's where I'm heading. So the first question is, how do I do the equivalent map-as-nondeterministic-apply when the strings is further wrapped in IO, as is the function being applied. getArgs >>= mapM_ (\s -> putStrLn s ) does double-duty of moving the argument from last place to the left, as it makes use of eta reduction. Because I have two things going on (list applicative and IO monad) I'm losing the slickness of using applicative syntax. Is there a nice way to make these work together? And more generally, how would you write such a construct? I'm naturally biased with my knowledge in other languages, so maybe there's a completely different "normal" way of approaching this? Thanks, -John

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How do I use Guards in record syntax?
by Dimitri DeFigueiredo 15 Apr '14

15 Apr '14

I'm having some trouble understanding where I can or cannot use guards inside record syntax. I'm writing a simple conversion routine, but I am not able to write it without inserting an extra let. Do I need a let expression here? Am I missing something? -------------- data OldTrade = OldTrade { oldprice :: Double , oldamount :: Double , oldbuysell :: String -- "buy", "sell" or "" } deriving( Eq, Show) data BuyOrSell = Buy | Sell | Unknown deriving(Eq, Show) data Trade = Trade { price :: Double , amount :: Double , buysell :: BuyOrSell } deriving( Eq, Show) convert :: OldTrade -> Trade convert ot = Trade { price = oldprice ot, amount = oldamount ot, buysell = let x | oldbuysell ot == "buy" = Buy | oldbuysell ot == "sell" = Sell | otherwise = Unknown in x } -- how do I eliminate the 'let' expression here? -- I wanted to write something like: -- -- buysell | oldbuysell ot == "buy" = Buy -- | oldbuysell ot == "sell" = Sell -- | otherwise = Unknown -------------- Thanks! Dimitri

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set multiplication
by Nishant 15 Apr '14

15 Apr '14

hi, I am trying to implement a set multiplication program in haskell. Input : [1,2,3] [4,5] [6,7] Ouput : [ [1,4,6] , [1,4,7] , [1,5,6] , [1,5,7] ...] I implemented it for constant number of inputs like setMul xs ys zs = [ [x] ++ [y] ++ [z] | x <- xs , y<-ys ,z <- zs] I am not able to generalize this for any number of lists. type signature would be : setMulMany :: [[a]] -> [[a]] Example : Input : [ [1,2,3] , [4,5] , [6,7]] Ouput : [ [1,4,6] , [1,4,7] , [1,5,6] , [1,5,7] ...] Regards. Nishant

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function not working as expected, type issues
by Alexej Magura 14 Apr '14

14 Apr '14

K, so I have a function that I’m writing that takes a *string* and a *count* and prints the *string* to STDOUT *count* times: displayD :: IO () -> String -> Int displayD string count = do (count > 0) && hPutStr stdout string (count > 0) && displayD string (count - 1) However, when I try to compile the file, I get several errors; here’s a pastebin: http://pastebin.com/DEsuAvfz What I’m trying to do here is check to see if *count* is greater than 0, and then if it is, then I’d like to print *string* to STDOUT until *count* equals 0. I tried doing it via pattern matching, like so: displayD string (count > 0) = do But I haven’t seen any examples of how to do pattern matching in functions using *do*, that is *IO*, so I’m unsure if the above is even anywhere near correct. Still very uncomfortable with declaring function types: I have a hard time figuring out which type is returned and what type is expected as an arg. My initial guess is that the *first* type specified is the one returned, but I’m not sure. -- Alexej Magura Sent with Airmail

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