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stack issues
by frantisek kocun 08 Jan '16

08 Jan '16

Hi, I'm new to cabal/stack. I want to use "stack build" to build my project. I can not use latest gkt >= 0.14.2 because of setup: The pkg-config package 'cairo' version >=1.2.0 is required but it could Neither can I now use older version gtk >=0.12.0 && < 0.13 While constructing the BuildPlan the following exceptions were encountered: -- Failure when adding dependencies: gtk: needed (>=0.12.0 && <0.13), 0.14.2 found (latest version available) needed for package: ships-0.1.0.0 What can i do? Thanks

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wxHaskell: add an external event from another thread
by Puck 08 Jan '16

08 Jan '16

Hello all, how is it possible to process external events, for example a succeeded getLine or a takeMVar with its values in the wxHaskell framework? The reason is, that I want to connect a erlang node to wxHaskell. I have found the functions Graphics.UI.WX.Events.newEvent :: String -> (w -> IO a) -> (w -> a -> IO ()) -> Event w a "Create a new event from a get and set function." Beside from, that I don't know what the get and set function shall do, I don't know, how I can add the new event to the event-loop of wx, and when it "fires". Do you know any possibilities? Thank you in advance

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Parametrizing [] as an instance of the Functor type class
by Olumide 08 Jan '16

08 Jan '16

According to LYH, the list is an instance of the Functor type class: instance Functor [] where fmap = map Why does 'f' not appear in this expression?, considering that Functor is defined as class Functor f where fmap :: (a -> b) -> f a -> f b Overall, I'm a bit confused about the relationship between the type constructor f and []. Thanks, - Olumide

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Pattern Matching for record syntax
by Jeon-Young Kang 07 Jan '16

07 Jan '16

Dear All. Hope your 2016 is off to a great start. I would like to get results from pattern matching or something. Here is the my code. data Person = Person {name :: String, age :: Int} names = ["tom", "sara"] -- list of names, String persons = [Person {name = "tom", age = 10}, Person {name="sara", age=9}, Person {name = "susan", age = 8}]. Is there any solution to get the age of "tom" and "sara"? I have no idea of pattern matching for this one. I've tried to use recursion, but I couldn't find any solution for list of records. Sincerely, Young

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Functor instance for ordered lists
by martin 04 Jan '16

04 Jan '16

Hello all, Data.List.Ordered is just a bunch of functions operating on ordinary Lists. Fmapping a function over an ordered list has the potential of blowing the ordering. Would it be possible to define a newtype for ordered lists where the order is guaranteed to be maintained? The functor instance then may have to re-order the elements. The problem I see is that data Ordlist a = ... would certainly require an Ord constraint on a, but where would I put it? I can put it on all the functions manipulating OrdLists, but I still wouldn't know how to define a functor instance, because a Functor a does not require Ord a.

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Tree-like Structure with unique elements on each level
by martin 04 Jan '16

04 Jan '16

Hello all, I recently modeled a tree like this: data Predicate a = Any | Pred (S.Set a) data Product a = Pany | Prod (S.Set(Predicate a, S.Set(Product a))) What I wanted to achieve was to have each element only once on each level. Hence the Sets. But this structure does not achieve this. I can easily duplicate elements on a level as long as their subordinates are different, so I might as well give up working with sets. Is there a way to construct a tree where each element can occur only once on each level?

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infinite type
by Julian Rohrhuber 04 Jan '16

04 Jan '16

The function g, when called with a binary function returns a type error which I’d really like to understand: why is this type “infinite" rather than just incomplete or something similar? I would have expected some kind of partial binding. Can someone help me with an explanation? Prelude> let f g = g . g Prelude> let sum x y = x + y Prelude> f sum <interactive>:14:3: Occurs check: cannot construct the infinite type: a ~ a -> a Expected type: (a -> a) -> a -> a Actual type: a -> a -> a Relevant bindings include it :: (a -> a) -> a -> a (bound at <interactive>:14:1) In the first argument of ‘f’, namely ‘sum’ In the expression: f sum With a similar call using lambda expressions, the error is different: Prelude> (\x y -> x + y) . (\x y -> x + y) <interactive>:32:1: Non type-variable argument in the constraint: Num (a -> a) (Use FlexibleContexts to permit this) When checking that ‘it’ has the inferred type it :: forall a. (Num a, Num (a -> a)) => a -> (a -> a) -> a -> a Prelude>

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Why must inflix constructors begin with a colon?
by Olumide 31 Dec '15

31 Dec '15

I'm new to Haskell and wondering why this restriction? For example (from LYH): infixr 5 :-: data List a = Empty | a :-: (List a) deriving (Show, Read, Eq, Ord) Thanks, - Olumide

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Doubts about functional programming paradigm
by Abhishek Kumar 29 Dec '15

29 Dec '15

I am a beginner in haskell.I have heard a lot about haskell being great for parallel programming and concurrency but couldn't understand why?Aren't iterative algorithms like MapReduce more suitable to run parallely?Also how immutable data structures add to speed?I'm having trouble understanding very philosophy of functional programming, how do we gain by writing everything as functions and pure code(without side effects)? Any links or references will be a great help. Thanks Abhishek Kumar

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How to show a predicate
by martin 29 Dec '15

29 Dec '15

Hello all, in my program, I do stuff with predicates (a->Bool). For the most part this representation is just fine, but at the very end I need to convert a resulting predicate into a String so I can write it to a file. Wenn I represent my predicates as Lists or Sets, then this is doable and I am tempted to do it this way. The only other option I could come up with was to have a representation of "everything", which would in my case be large (10^8) but finite. Then I could construct a List or a Set at the very end, as [x | x<-everything, p x] without having explicit sets in the intermediate steps. I cannot see any other option, but I thought I better ask.

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