August 2007 - Haskell-Cafe

Slow IO or bad code?
by Vimal 09 Aug '07

09 Aug '07

Hi I am practicing writing code in haskell, by solving problems at this site. http://spoj.pl. The problem http://spoj.pl/problems/FASHION , is pretty simple. 1. Given two lists A,B , of N numbers, sort them and take sum of products. i.e. Sum ai * bi I wrote a code, but seems to give "Time limit exceeded"! >> Beginning of CODE loop t function | t == 1 = do function | otherwise = do { function; loop (t - 1) function } prod [] [] = 0 prod (a:as) (b:bs) = a*b + prod as bs to_int :: [String] -> [Integer] to_int [] = [] to_int (x:xs) = (read x) : to_int xs doit = do n <- getLine a <- getLine b <- getLine let la = to_int (words a); lb = to_int (words b); in print (prod la lb) main = do t <- getLine loop (read t) doit << END OF CODE I would love to see if there is any improvement that can be done, to the code ... Thanks! Vimal

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Operator overloading
by rodrigo.bonifacio 09 Aug '07

09 Aug '07

Hi all. I want to overload the operator "^" for working instead of the following "+++" operator: (+++) :: String -> [[String]] -> [[String]] x +++ y = [ x:e | e<-y ] How can I overload the "^" operator? Thanks a lot. Rodrigo.

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(no subject)
by Thomas Hartman 09 Aug '07

09 Aug '07

In the following code which uses template haskell, how can I get back the macro-expanded code generated from $(inferStartState ''MyState) I *can* recover the macro-expanded code for $(cnst 1 "x") using a debugging technique bulat describes on his tutorial at http://www.haskell.org/bz/th3.htm You can see what's going on in the function debugTemplates below. I'm trying to do this actually, to better understand how HAppS deals with state. It's a bit opaque now since the example on the tutorial uses TH. I think I would understand it better if I had code that didn't depend on TH. (MyState is from the happs tutorial at http://www.haskell.org/haskellwiki/HAppS_tutorial ) thanks! thomas. {-# OPTIONS -fglasgow-exts -fth #-} module MyState where import HAppS import HAppS.Protocols.SimpleHTTP2 import Data.Monoid import Data.Typeable import Control.Monad.State (get, put) import Language.Haskell.TH import Language.Haskell.TH.Syntax data MyState = MySt { appVal :: Int } deriving (Read, Show, Typeable) instance Serialize MyState where encodeStringM = defaultEncodeStringM decodeStringM = defaultDecodeStringM instance Monoid MyState where mempty = MySt 0 mappend (MySt x) (MySt y) = MySt (x+y) -- in ghci... -fth, :m + -- ghci... :t (inferStartState ''MyState) :: (Quasi m) => m [Dec] $(inferStartState ''MyState) -- boilerplate that will eventually be SYB -- ghci... :t cnst 1 "x" :: (Monad m) => m Exp cnst n s = return (LamE (replicate n WildP) (LitE (StringL s))) dumpSplice splice = runQ splice >>= putStrLn . pprint debugTemplates = do dumpSplice (cnst 1 "x") dumpSplice (inferStartState ''MyState) {- *MyState> debugTemplates \_ -> "x" Template Haskell error: Can't do `reify' in the IO monad *** Exception: user error (Template Haskell failure) -} --- This e-mail may contain confidential and/or privileged information. If you are not the intended recipient (or have received this e-mail in error) please notify the sender immediately and destroy this e-mail. Any unauthorized copying, disclosure or distribution of the material in this e-mail is strictly forbidden.

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New Eq instance
by rodrigo.bonifacio 09 Aug '07

09 Aug '07

Hello, I had defined the follwing data type: data Step = Step Id Scenario Action State Response How can I define Step as an "Eq Instance", in such way that two steps are equals if they have the same Id (Id is defined as a synonimous for the String type). I tried the following code, but something is wrong.... instance Eq Step where Step id1 scenario1 action1 state1 response1 == Step id2 scenario2 action2 state2 response2 = id == id _ == _ = False ps.: sorry, this kind of basic question can be sent to this list? Thanks in advance. Rodrigo.

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Using haskell with emacs
by Paulo J. Matos 09 Aug '07

09 Aug '07

Hi all, I'm starting to learn haskell by my own, being currently mostly a Common Lisp, Scheme, C++ programmer... I've got the haskell emacs mode but can't find a manual. Moreover, I've found some keybindings on the net but nothing that allows me to start an interpreter in emacs and send definitions, one by one to the interpreter. Is this possible? Is there any good reference of the emacs keybindings for haskell mode? Cheers, -- Paulo Jorge Matos - pocm at soton.ac.uk http://www.personal.soton.ac.uk/pocm PhD Student @ ECS University of Southampton, UK

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Using HList Records
by Rahul Kapoor 09 Aug '07

09 Aug '07

Using The Darcs version of HList (http://darcs.haskell.org/HList/) I can do simple things with (H)lists just fine, so --angus = Key 42 -- .*. Name "Angus" -- .*. Cow -- .*. Price 75.5 -- .*. HNil --main = putStrLn (show angus) works, On the other hand examples from the HList paper for extensible records like --key = firstLabel FootNMouth "key" --name = nextLabel key "name" --breed = nextLabel name "breed" --price = nextLabel breed "price" do not compile with errors --Main.hs:15:8-17: Not in scope: `firstLabel' --Main.hs:16:8-16: Not in scope: `nextLabel' A quick show "ghc-pkg describe HList" shows that --exposed-modules: HList --hidden-modules: Label4 CommonMain Variant GhcSyntax GhcRecord -- Record HZip TIC TIP HTypeIndexed HOccurs HArray GhcExperiments -- HListPrelude TypeEqBoolGeneric TypeEqGeneric1 TypeCastGeneric1 -- FakePrelude So the question is do I need to play around with the cabal file to expose the right stuff? Or has the syntax changed? I get errors running "./setup haddock" (configure, build, install, are fine). Are the HList api docs available anywhere online? Rahul

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Re: default for quotRem in terms of divMod?
by Christian Maeder 08 Aug '07

08 Aug '07

Isaac Dupree wrote: > In class Integral, divMod has a default in terms of quotRem. > (quot,rem,div,mod all have defaults as the both-function they're part > of.) I'm sure divMod is more natural than quotRem to implement for some > types... so why doesn't quotRem have a default in terms of divMod? it > has no default! Then the "minimal to implement" will change from > (toInteger and quotRem) to (toInteger and (quotRem or divMod)). > > Isaac while I don't care if quotRem or divMod should be implemented. I oppose to give both default implementations in terms of the other. Already for the class Eq either == or /= must be defined, with the unpleasant effect that an empty instance like: instance Eq T leads to a loop (when == or /= is called on elements of type T). The empty instance does not even raise a warning about unimplemented methods (since the default definition is used). I'd rather prefer to remove /= as method of Eq. Cheers Christian

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Defining new operators
by rodrigo.bonifacio 08 Aug '07

08 Aug '07

Hello, I have created the following function: dist :: String -> [[String]] -> [[String]] dist x y = [ x:e | e<-y ] eg.: dist "1M" [[], ["2M"], ["2M, 3M"]] = [["1M"],["1M","2M"],["1M","2M, 3M"]] How can I create an operator that perform the same function as dist? I want to write something like: "1M" ++ [[], ["2M"], ["2M", "3M"]] Thanks in advance. Rodrigo.

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Newbie question (again!) about phantom types
by peterv 08 Aug '07

08 Aug '07

I’m having difficulty to understand what phantom types are good for. Is this just for improving runtime performance? I read the wiki, and it says "this is useful if you want to increase the type-safety of your code", but the code below does not give a compiler error for the function test1, I get a runtime error, just like test2. Thanks, Peter -- CODE -- -- With phantom types data T1 a = TI1 Int | TS1 String deriving Show foo1 :: T1 String -> T1 String -> T1 String foo1 (TS1 x) (TS1 y) = TS1 (x++y) test1 = foo1 (TI1 1) (TI1 2) -- Shouldn't this give a compiler error instead of a runtime error? -- Without phantom types data T2 = TI2 Int | TS2 String deriving Show foo2 :: T2 -> T2 -> T2 foo2 (TS2 x) (TS2 y) = TS2 (x++y) test2 = foo2 (TI2 1) (TI2 2) No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.5.476 / Virus Database: 269.11.8/941 - Release Date: 07/08/2007 16:06

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mutually recursive types
by rodrigo.bonifacio 08 Aug '07

08 Aug '07

Hi, I am learning the haskell programming language and had tried to define the following types: type Scenario = (String, String, [Step]) type Step = (String, Scenario, String, String, String) Notice that Scenario depends on a list of steps and Step has a dependence with scenario. I know that this is a kind of "bad smell".... in Haskell, are there any pattern or language idiom to deal with cyclical dependences? Regards, Rodrigo.

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