July 2011 - Beginners

Cannot install cairo with cabal...
by Sean Charles 11 Aug '11

11 Aug '11

flushed as in happy, not like a buffer, with my recent OpenGL and haskell hacking I attempted to install cairo to try rendering a cairo generated image into a GL texture but once again, the sometimes obstreperous cabal is making it hard. <rant> I sometimes think that cabal is the "weak" point for newbies like me accepting Haskell in all its goodness... if you can't cleanly and simply install libraries then ultimately you will walk away for something less obstructive... </rant> Can anybody tell me that this means and how to solve it ? Thanks, Sean. ----- scharles@ubuntu:~/stuff/hackinge/objitsu/trunk/haskell/felt/src$ cabal --version cabal-install version 0.10.2 using version 1.10.2.0 of the Cabal library scharles@ubuntu:~/stuff/hackinge/objitsu/trunk/haskell/felt/src$ ghc --version The Glorious Glasgow Haskell Compilation System, version 6.12.3 scharles@ubuntu:~/stuff/hackinge/objitsu/trunk/haskell/felt/src$ ghci --version The Glorious Glasgow Haskell Compilation System, version 6.12.3 ----- /tmp/cairo-0.12.013023/cairo-0.12.0/Gtk2HsSetup.hs:25:2: warning: #warning Setup.hs is guessing the version of Cabal. If compilation of Setup.hs fails use -DCABAL_VERSION_MINOR=x for Cabal version 1.x.0 when building (prefixed by --ghc-option= when using the 'cabal' command) [1 of 2] Compiling Gtk2HsSetup ( /tmp/cairo-0.12.013023/cairo-0.12.0/Gtk2HsSetup.hs, /tmp/cairo-0.12.013023/cairo-0.12.0/dist/setup/Gtk2HsSetup.o ) /tmp/cairo-0.12.013023/cairo-0.12.0/Gtk2HsSetup.hs:201:69: Couldn't match expected type `PackageDBStack' against inferred type `PackageDB' In the sixth argument of `registerPackage', namely `packageDb' In the expression: registerPackage verbosity installedPkgInfo pkg lbi inplace packageDb In a case alternative: _ | modeGenerateRegFile -> die "Generate Reg File not supported" | modeGenerateRegScript -> die "Generate Reg Script not supported" | otherwise -> registerPackage verbosity installedPkgInfo pkg lbi inplace packageDb cabal: Error: some packages failed to install: cairo-0.12.0 failed during the configure step. The exception was: ExitFailure 1

4 10

Network client - reading and writing to a socket
by Manfred Lotz 02 Aug '11

02 Aug '11

Hi there, I'm trying to write a network client which connects to an IMAP server thus reading from and writing to socket 143. I have an initial solution which works except that laziness bites me. Perhaps the whole solution using connectTo and handle is the wrong approach. Here is the code: module Main where import System.IO import qualified Data.ByteString.Char8 as B import Network server :: String server = "127.0.0.1" port :: PortID port = PortNumber 143 sendMsg :: Handle -> String -> IO () sendMsg h m = do putStrLn ("Sending command: " ++ m) B.hPutStrLn h (B.pack m) >> hFlush h recvMsg :: Handle -> IO B.ByteString recvMsg h = do c <- B.hGet h 1 c' <- B.hGetNonBlocking h 40000 return $ B.concat [c,c'] main :: IO () main = withSocketsDo $ do h <- connectTo server port hSetBuffering h LineBuffering imapDialog h imapDialog :: Handle -> IO () imapDialog h = do greet <- recvMsg h B.putStrLn greet sendMsg h "1 LOGIN manfred \"password\"" resp <- recvMsg h B.putStrLn resp sendMsg h "2 SELECT HAM-learn" resp1 <- recvMsg h B.putStrLn resp1 sendMsg h "3 FETCH 1:* (uid flags internaldate body[header.fields (Message-Id)])" resp2 <- recvMsg h B.putStrLn resp2 sendMsg h "4 FETCH 1 (rfc822)" resp3 <- recvMsg h B.putStrLn resp3 The problem is that the message itself is some 30K big and I only get some 16K of the message. How could I force to get the whole message? -- Manfred

5 14

Problem with catching error
by Alexey G 01 Aug '11

01 Aug '11

Hello. I have some problem with catching error. My function try to read value from string and if it's fails - return default (d). With this: >import Control.Exception >readIOWith :: Read a => a -> String -> IO a >readIOWith d x = catch tryRead errHandler > where errHandler :: SomeException -> IO a > errHandler _ = return d > tryRead = readIO x I have error: Could not deduce (a ~ a1) from the context (Read a) bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:(19,1)-(22,33) `a' is a rigid type variable bound by the type signature for readIOWith :: Read a => a -> String -> IO a at BWClub/Common/Helpers/Packets.hs:19:1 `a1' is a rigid type variable bound by the type signature for errHandler :: SomeException -> IO a1 at BWClub/Common/Helpers/Packets.hs:21:11 In the first argument of `return', namely `d' In the expression: return d In an equation for `errHandler': errHandler _ = return d But this code works: >readIOWith d x = catch tryRead (\e -> print (e :: SomeException) >> return d) > where tryRead = readIO x Sorry for my english.

3 3

Haskell-style types in C or C++
by Christopher Howard 31 Jul '11

31 Jul '11

One of the things that I love about Haskell is the syntax for creating user defined types. E.g.: Data QueryResult = NoResult | DatabaseError String | Results [String] I can prototype an entire program around these self-made types and it is a lot of fun. Out of intellect curiosity, though: what would be the equivalent construction in C or C++? (Say, for the type defined above). -- frigidcode.com theologia.indicium.us

7 7

No instance for (Show a)
by Ovidiu Deac 31 Jul '11

31 Jul '11

For some reason ghc complains about not being able to call show on an Integer (?!?!?) Please enlighten me! ovidiu This is the hspec: it "shows one element" ( show (push 1 EmptyStack) ≡ "Stack(1)") ...this is the code: data Stack a = EmptyStack | StackEntry a (Stack a) deriving(Eq) instance Show (Stack a) where show s = "Stack(" ⊕ (showImpl s) ⊕ ")" where showImpl EmptyStack = "" showImpl (StackEntry x _) = show x ...and this is the error: src/Stack.hs:12:22: No instance for (Show a) arising from a use of `showImpl' In the first argument of `(++)', namely `(showImpl s)' In the second argument of `(++)', namely `(showImpl s) ++ ")"' In the expression: "Stack(" ++ (showImpl s) ++ ")"

2 1

Ambiguous type variable
by Ovidiu Deac 30 Jul '11

30 Jul '11

I'm playing with Haskell so I wrote a stack module (see the code below). I have a problem with the pop function which returns a tuple (Nothing, EmptyStack) if called with an EmptyStack. I kind of understand that the compiler cannot cannot figure out what type to use for a. But how could I tell the compiler that if the list is empty I don't care about that type? Thanks, Ovidiu ///////////////// This is the hspec ... it "pop empty stack gives Nothing" ( (pop EmptyStack) ≡ (Nothing, EmptyStack)) ... This is the code: module Stack where import Prelude data Stack a = EmptyStack | StackEntry a (Stack a) deriving(Show, Eq) ... pop :: Stack a → (Maybe a, Stack a) pop EmptyStack = (Nothing, EmptyStack) pop (StackEntry a s) = ((Just a), s) ...and this is the error I get: test/TestStack.hs:20:28: Ambiguous type variable `a0' in the constraint: (Eq a0) arising from a use of `==' Probable fix: add a type signature that fixes these type variable(s) In the second argument of `it', namely `((pop EmptyStack) == (Nothing, EmptyStack))' In the expression: it "pop empty stack gives Nothing" ((pop EmptyStack) == (Nothing, EmptyStack)) In the second argument of `describe', namely `[it "empty stack is empty" (isEmpty EmptyStack), it "non-empty stack is not empty" (not (isEmpty (push 10 EmptyStack))), it "push then pop retrieves the same value" ((pop $ push 10 EmptyStack) == (Just 10, EmptyStack)), it "push push then pop retrieves the last value" ((pop $ push 2 (push 1 EmptyStack)) == (Just 2, (push 1 EmptyStack))), ....]' make: *** [test] Error 1

3 5

Re: [Haskell-beginners] minimal Haskell concepts subset
by Davi Santos 30 Jul '11

30 Jul '11

Hi Matt, the first and best I found was http://learnyouahaskell.com/chapters. The point is that I spent too much time (almost a year) learning, and I am not -productive- yet. Productive in the sense of time writing and debugging, not in the sense of number of characters written. Nothing in particular, it just was too much time. My background is, in order of experience: Java, C++, Delphi, Python, C#, Bash, Scala, Basic, Pascal and maybe others I forgot. The first two I really know and could learn the best practices. The rest I know only the minimal concepts but could do something important already. Haskell has been harder than Basic, my first programming language. When I was 13-year old, somebody told me: "each line is an order the computer follows in sequence". That was enough intuitive to me. After that I was taught: for, goto, if, print and input. The last concepts and harder were the while and the subroutines. Maybe I am get messy with the lots of alternatives to do a same thing in Haskell. After months in Java I started the habit of always search for the best practices - and I found them. Ok, it is long enough email. Maybe I have some kind of non-haskell-related addiction. Davi

3 3

Building Haskell Platform with shared libraries?
by Mister Globules 30 Jul '11

30 Jul '11

Hi, I'm installing the Haskell Platform (2011.2.0.1) from source on Kubuntu. If I follow the recommended steps (configure, make, make install) it doesn't build any shared libraries. I've tried a few hacks (e.g. export EXTRA_CONFIGURE_OPTS=--enable-shared), but they result in build errors. Currently, I do a "cabal --reinstall install" for each of the packages in the platform, but I was wondering if there's a more "official" way to do it when building the platform itself. - Globules

1 0

question on types
by Jake Penton 29 Jul '11

29 Jul '11

Yikes. I have been doing a fair bit of productive programming in Haskell, thinking that I am making a bit of progress. Then I hit something that is apparently *really simple* that I do not understand at all. How discouraging. Here is the code that makes me realize I don't understand types or type inference very well at all yet: f :: a f = 1 When I try to load the above, ghci gives me: No instance for (Num a) arising from the literal `2' In the expression: 2 In an equation for `f': f = 2 Failed, modules loaded: none. Ok, so then I try: g:: (Num a) => a g = 2 This compiles. Why? I mean, why is the first example (defining f) wrong, but the second example (defining g) ok? A slight variation on this is: h:: a h = 'a' to which ghci replies: Couldn't match type `a' with `Char' `a' is a rigid type variable bound by the type signature for c :: a at /Users/David/Project/EoP/ch04/weak.hs:114:1 In the expression: 'a' In an equation for `c': c = 'a' This last example is probably the most basic one which I need to understand. But, why is the problem apparently a different one than in the definition of "f" above? Of course, I cannot think of a reason to actually define things as shown above under ordinary circumstances. The code above is just boiled down to the simplest case I could find to illustrate my confusion. I guess I interpret "f::a" to mean "f is some (any) type a". So why can't it be whatever "1" is, which I suppose is Integer. What is the type system looking for? And why does the constraint (Num a) make things ok? Please point me in the direction of any reading I should do to clear up my confusion. TIA. - Jake -

8 19

Re: [Haskell-beginners] lazyness and tail recursion
by Ovidiu Deac 29 Jul '11

29 Jul '11

Now I think I understand. Thanks for the explanation! ovidiu On Fri, Jul 29, 2011 at 12:04 AM, Will Ness <will_n48(a)yahoo.com> wrote: > ... the start of the function, so to speak. > > If we have (f x = a : f b), the distance between this entry into f, and the > next one, has constant length: (a :). If we have (f x = "123" ++ f y), it's > constant too, just bigger than 1 - no problem, it will all get consumed in > constant number of steps, there's no explosion in number of operations. > > But if we have (f x = f a ++ f b) e.g., then (f a) will add some, and its > calls will get some, and it all grows out of control - and all this needs to > be maintained by the system, to get to the final (f b). Not nice. > > But anyway I'm no expert; it's just how I explain it to myself, from > Prolog's "tail recursion modulo cons" perspective. We don't need to wait for > a function to return to prepend something onto its result; we prepent this > to a yet-to-be-filled list, and pass that stump into the tail-recursive > function. Presto, all it fine. Same in Haskell, no second function call will > get actually called until what's before it gets consumed - so there's no > proble if that's constant in size. But I repeat myself. :) > > Cheers, > Will > > > --- On *Thu, 7/28/11, Ovidiu Deac <ovidiudeac(a)gmail.com>* wrote: > > > From: Ovidiu Deac <ovidiudeac(a)gmail.com> > Subject: Re: [Haskell-beginners] lazyness and tail recursion > To: "Will Ness" <will_n48(a)yahoo.com> > Cc: beginners(a)haskell.org > Date: Thursday, July 28, 2011, 10:27 AM > > > If I got it right, the idea is to keep the number of operations > between the recursive call and the end of the function constant. > > On Wed, Jul 27, 2011 at 1:38 AM, Will Ness <will_n48(a)yahoo.com<http://us.mc1145.mail.yahoo.com/mc/compose?to=will_n48@yahoo.com>> > wrote: > > Ovidiu Deac <ovidiudeac <at> gmail.com> writes: > > > >> > >> I've read the paragraph below from Real World Haskell > >> > > ( > http://book.realworldhaskell.org/read/efficient-file-processing-regular-exp… > >> section "An important aside: writing lazy functions") > >> > >> I can get a feeling why lazyness compensates for the lack of tail > >> recursion but I don't really understand why it would work in general. > >> > >> My understanding is that that a function which returns a list with > >> non-tail recursion would only work this way with certain usage > >> patterns. In this case the caller of the funtion will have to use the > >> string in a stream-like fashion. > > > > The concept of tail recursion modulo cons is similar. Wikipedia has an > article. > > Basically it's about keeping one execution frame above the tail-recursive > one. > > Lazily accessing a list from the top is a lot like adding elements to it > one by > > one at the bottom. Prefixing a recursive call with an operation of fixed > number > > of operations is OK, because they will get consumed and the recursive > case > > called, in a fixed number of steps - the instance between current point > and next > > recursive invocation point won't grow, only get increased at times by a > set > > amount and then get consumed. > > > > The problem with non-tail recursion in lazy setting is that the distance > grows > > between the current point and the next invocation, and so the amount of > "what to > > do next?" information grows all the time. In imperative setting it gets > flipped > > into "what to do after the invocation" but it still grows. In > tail-recursion > > (even modulo cons, or (++) with fixed prefix) it's bounded, finite, > constant. > > Looking at Scheme example code in WP "continuation-passing style" article > might > > help. There we see as in translations of tail-recursive functions the > > constructed continuation does not grow in unlimited fashion, instead only > maybe > > getting prefixed by a fixed amount of operations at times. > > > > Does it make any sense? :) > > > > > > _______________________________________________ > > Beginners mailing list > > Beginners(a)haskell.org<http://us.mc1145.mail.yahoo.com/mc/compose?to=Beginners@haskell.org> > > http://www.haskell.org/mailman/listinfo/beginners > > > >

1 0