- Beginners - Haskell.org

When did Haskell become fun for you?
by aditya siram 05 Jul '09

05 Jul '09

Hi all, I have been pursuing Haskell for about a year now and the community has been awesome. The learning curve has been ( and is still ) steep for me because of the sheer density of the concepts, but it has changed the way I think about programming. Relating it to music, Haskell is like jazz, the concepts are heady and the entry into that world requires a high level of patience and discipline. But in the end, you want to zone out and play. How long was it before you could just play Haskell? Thanks ... -deech

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Re: [Haskell-beginners] help with types and composition
by Troy Pracy 04 Jul '09

04 Jul '09

I've just started learning Haskell and I've been wondering about this issue as well. I can usually work out a point-free version by carefullty deriving it step-by-step, but I was wondering if Haskell had composition operators/functions for dealing with the various forms of composition where a 2-arg function is involved. I've played around with J (APL's successor) a little and noticed that J has various options for composing two functions (Ponit-free, or "implicit" style is very important in J). Some of the distinctions have to do with J's native array operations and aren't relevant here, but many are. Here are Haskell versions... (note: "monadic" below isn't used in the Haskell/CT sense - "monadic" and "dyadic" in J jsut refer to how many arguments an operator acts on) -- hook is the J dyadic hook as a function. hook :: (a->b->c) -> (d->b) -> a -> d -> c hook f g = \x y -> f x (g y) -- J's monadic hook mhook :: (a->b->c) -> (a->b) -> a -> c mhook f g = \x -> (hook f g) x x -- J's monadic fork fork :: (a->b->c) -> (d->a) -> (d->b) -> d -> c fork f g h = \x -> f (g x) (h x) -- J's dyadic fork dyfork :: (a->b->c) -> (d->e->a) -> (d->e->b) -> d -> e -> c dyfork f g h = \x y -> f (g x y) (h x y) -- J's dyadic @ or @: - composition of 1-arg fn with 2-arg fn compose12 :: (a->b) -> (c->d->a) -> c -> d -> b compose12 f g = \x y -> f (g x y) (@:) = compose12 {- J's dyadic & or &: - composition of 2-arg fn with 1-arg fn, resulting in a 2-arg fn (f&:g) which applies g to *both* args before passing them to f. Haskell's composition operator and partial application allow a composition of such fns (f . g) where g is applied only to the first arg. -} compose21 :: (a->a->b) -> (c->a) -> c -> c -> b compose21 f g = \x y -> f (g x) (g y) (&:) = compose21 / /I know a lot of Haskell is written in point-free style and I would have thought Haskell would have operators for some of this, but judging from the previous responses, it looks like it might not. That surprises me, since some of this seems to crop up a lot, but as I said I've just started learning Haskell, so I guess I'll have to give myself some time to absorb the Haskell way of doing things.

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GHC compiling for different platforms
by Bernhard Lehnert 04 Jul '09

04 Jul '09

Hi folks! When I compile a program under windows - will the .exe run under any windows? Do I have to distribute only the .exe or more files? When I compile a program under a 64bit-Linux on an AMD-CPU, will the executable run under 32bit and/or Intel-CPU as well? Are there any, say, compiler flags, to improve portability? Thanks, Bernhard

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help with types and composition
by Britt Anderson 03 Jul '09

03 Jul '09

I would like to write a function that takes two lists of numbers and returns a number. For simplicity, let's assume an integer, i.e. myfunc:: [Integer]->[Integer]->Integer But I can't quite get things right when trying to define myfunc = ... and making the arguments implicit. My difficulty might be seen with the following two functions, let, f1 = (zipWith ($)) . (map (*)) f2 = sum if I set the types such that f1 returns [Integer] and sum accepts [Integer] it seems that somehow I should be able to compose them in to a single function that takes two lists and returns the number, but f1 . f2 and numerous variations don't work. I can curry some list l and f1 . (f2 l) works, but is there a way to get f1 and f2 combined into a function that accepts two lists? My problem is not really about writing the function but about composition. I can write the function with arguments or a local lambda that suits my practical purpose. But the fact that I think I should be able to compose f1 . f2 somehow reveals a conceptual misunderstanding that I hope someone will help me clear up. Thank you, Britt Anderson

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Fwd: [Haskell-beginners] help with types and composition
by Alexander Dunlap 03 Jul '09

03 Jul '09

Oops...wrong list address...sorry ---------- Forwarded message ---------- From: Alexander Dunlap <alexander.dunlap(a)gmail.com> Date: Thu, Jul 2, 2009 at 8:54 PM Subject: Re: [Haskell-beginners] help with types and composition To: Britt Anderson <britt(a)uwaterloo.ca>, haskell-beginners(a)haskell.org On Thu, Jul 2, 2009 at 10:54 AM, Britt Anderson<britt(a)uwaterloo.ca> wrote: > I would like to write a function that takes two lists of numbers and returns > a number. For simplicity, let's assume an integer, i.e. > myfunc:: [Integer]->[Integer]->Integer > But I can't quite get things right when trying to > define myfunc = ... > and making the arguments implicit. > > My difficulty might be seen with the following two functions, > let, > > f1 = (zipWith ($)) . (map (*)) > > f2 = sum > > > if I set the types such that f1 returns [Integer] and sum accepts [Integer] > it seems that somehow I should be able to compose them in to a single > function that takes two lists and returns the number, but f1 . f2 and > numerous variations don't work. I can curry some list l and f1 . (f2 l) > works, but is there a way to get f1 and f2 combined into a function that > accepts two lists? > > > My problem is not really about writing the function but about composition. I > can write the function with arguments or a local lambda that suits my > practical purpose. But the fact that I think I should be able to compose f1 > . f2 somehow reveals a conceptual misunderstanding that I hope someone will > help me clear up. > > Thank you, > Britt Anderson > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > Try (f2 .) . f1 . The type of (f2 .) is (a -> [Integer]) -> a -> Integer and the type of (.) is (b -> c) -> (a -> b) -> (a -> c) so for the second (.), b :: ([Integer] -> [Integer]), c :: ([Integer] -> Integer) and a :: [Integer] so the final type is [Integer] -> [Integer] -> Integer. (It might be easier to work out the types yourself if you can - my description is doubtless rather difficult to follow.) A word of caution, though - most Haskellers in my experience prefer to not do this type of composition; they find it clearer to specify one or both of the arguments explicitly. Hope that helps you, Alex

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Problems with IO actions
by Adam Bergmark 02 Jul '09

02 Jul '09

I have an issue with IO actions. At the moment I have the following snippet of code, which works as I expect it to. mapM_ (writeChan ch) . lines =<< hGetContents h That is, it gets each line in the file and writes it to the channel. Previously I had just: writeChan ch =<< hGetContents h but when i readChan this I get the empty string every time. I have two questions: * I'm still a little unsure of why mapM_, sequence and the like are necessary, could someone please explain this, or point me somewhere where I can read up on it? * Why does this problem occur when there is only one action to perform? Cheers / Adam Bergmark

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Is there a replacement for network-minihttp that allows lazy downloads?
by Adam Bergmark 02 Jul '09

02 Jul '09

I'm trying to write a program that will download possibly large files using HTTP. In Real World Haskell I read: Warning The HTTP library used here does not read the HTTP result lazily. As a result, it can result in the consumption of a large amount of RAM when downloading large files such as podcasts. Other libraries are available that do not have this limitation. We used this one because it is stable, easy to install, and reasonably easy to use. We suggest mini-http, available from Hackage, for serious HTTP needs. Looking at http://hackage.haskell.org/cgi-bin/hackage-scripts/package/network-minihttp I see that minihttp is deprecated. Is there a replacement I can use? / Adam Bergmark

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ghc profiler beginner
by José Prous 30 Jun '09

30 Jun '09

I never used the ghc profiler before, when I try to compile the program with: ghc -prof -auto -O2 -o Main --make Main.hs I get: Could not find module `Data.Binary.IEEE754': Perhaps you haven't installed the profiling libraries for package `data-binary-ieee754-0.3'? Use -v to see a list of the files searched for. Is there a way to tell the profiler to ignore the external libraries or I have to recompile all the libraries with profiling enabled? thanks

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High precision doubles
by Aaron MacDonald 29 Jun '09

29 Jun '09

Is there anything in Haskell that can represent doubles with higher precision than the regular kind? I'm working with formulas that generate sets of values, and values that I know should be zero are ending up as things like -7.105427357601002e-15 or -1.7763568394002505e-14, which is causing all kinds of butterfly effects. I've heard of Rational, but I need to also work with irrational numbers (namely, pi and anything that comes from cos and sin). What other options do I have? - Aaron

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Rigid type variables match error
by Darryn 28 Jun '09

28 Jun '09

Hi, I wonder if anyone can explain what is going on here and what to do about it. I'm fairly new to Haskell, so apologies in advance if my question seems naive. I've cut my code down to a minimum that reproduces the problem. I receive an error in GHCI in the following code, complaining that it cannot match the rigid type variables for the instance definition for Ainst for the function a3. Can anyone advise about what to do about it? ------------------------------------------------ class A a where a1 :: a a2 :: a -> a a3 :: (B b) => b -> a class B b where b1 :: Int -> b data (B b) => Ainst b = I | J (Ainst b) | K b instance (B b) => A (Ainst b) where a1 = I -- a2 :: (B b, A a) => a -> a a2 = J -- a3 :: (B b, A a) => b -> a a3 = K -- Error! -- a3 = K `asTypeOf` a3 -- Error even with this! data Binst = Val Int instance B Binst where b1 = Val ------------------------------------------------ Test5.hs:17:9: Couldn't match expected type `b1' against inferred type `b' `b1' is a rigid type variable bound by the instance declaration at Test5.hs:12:12 `b' is a rigid type variable bound by the type signature for `a3' at Test5.hs:5:13 Expected type: b -> Ainst b1 Inferred type: b -> Ainst b In the expression: K `asTypeOf` a3 In the definition of `a3': a3 = K `asTypeOf` a3 Failed, modules loaded: none.

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