June 2009 - Haskell-Cafe

test-framework success
by Simon Michael 28 Jun '09

28 Jun '09

Max - I was thinking about how to drive my new shell tests with your framework. I have: data ShellTest = ShellTest { filename ∷ String ,command ∷ String ,stdin ∷ Maybe String ,stdoutExpected ∷ Maybe String ,stderrExpected ∷ Maybe String ,exitCodeExpected ∷ Maybe ExitCode } deriving (Show) parseShellTest ∷ FilePath → IO ShellTest runShellTest ∷ ShellTest → IO Bool I converted to hunit tests with shellTestToHUnitTest t = filename t ~: do {r ← runShellTest t; assertBool "" r} and tied test-framework into my test runner with main = do args ← getArgs let (opts,files) = partition ((=="-").take 1) args hunittests ← mapM (λf → parseShellTest f >>= (return ∘ shellTestToHUnitTest)) files withArgs opts $ defaultMain $ concatMap hUnitTestToTests hunittests With surprisingly little code and pain, the new runner runs my 14 shell tests, with colored output, in the same time as the old one - 3.5s on this macbook. But, I can now add -j8 and get the same results output in.. 0.13s. This quite surprised me, and now I want to say: thank you very much! :) Best, -Simon $ time tools/shelltest2 tests/*.test -j8 :tests/eliding-balance.test: [OK] :tests/missing-real-and-virtual-amt.test: [OK] :tests/null-accountname-component.test: [OK] :tests/parens-in-account-name.test: [OK] :tests/sample-balance-depth.test: [OK] :tests/sample-balance-o.test: [OK] :tests/sample-balance.test: [OK] :tests/unbalanced.test: [OK] :tests/unicode-account-matching.test: [OK] :tests/unicode-balance.test: [OK] :tests/unicode-description-matching.test: [OK] :tests/unicode-print.test: [OK] :tests/unicode-register.test: [OK] :tests/virtual.test: [OK] Test Cases Total Passed 14 14 Failed 0 0 Total 14 14 real 0m0.132s user 0m0.104s sys 0m0.104s

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What is an "expected type" ...
by michael rice 28 Jun '09

28 Jun '09

as opposed to an "inferred type"? Michael

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"shadowing" keywords like "otherwise"
by Vasili I. Galchin 28 Jun '09

28 Jun '09

Hello, I am currently using ghc 6.8.2. With the following, swishParse :: String -> String -> SwishStateIO (Maybe RDFGraph) swishParse fnam inp = do { fmt <- gets $ format ; case fmt of N3 -> swishParseN3 fnam inp otherwise -> do { swishError ("Unsupported file format: "++(show fmt)) 4 ; return Nothing } } I am receiving a shadow warning: Swish/HaskellRDF/SwishCommands.hs:304:12: Warning: Defined but not used: `otherwise' It seems to me that in the code base somewhere that there is a "redefine" of the keyword"otherwise". I haven't read the Haskell 98 Report but I thought that it was not possible to redefine keywords. ?? Kind regards, Vasili

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A Reader Monad Tutorial
by Henry Laxen 28 Jun '09

28 Jun '09

Dear Group, If any of you are struggling with understanding monads, I've tried to put together a pretty through explanation of what is behind the Reader monad. If you're interested, have a look at: http://www.maztravel.com/haskell/readerMonad.html Enjoy. Henry Laxen

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Re: [Haskell-cafe] Rigid type variable error
by Darryn 28 Jun '09

28 Jun '09

Thanks for the help previously received, but I still cannot seem to get on top of this. The types for the constructor K will not resolve and I'm at a loss to work out what to do with it. If anyone can offer a further explanation and help I would be very grateful. My code (File Test5.hs): ---------------------------- {-# LANGUAGE ExistentialQuantification #-} class A a where a1 :: a a2 :: a -> a a3 :: (B b) => b -> a class B b where b1 :: Int -> b --data Ainst b = I | J (Ainst b) | K b -- a3 :: (B b, A a) => b -> a -- yet without the constraint on K, K :: b -> Ainst b -- so the above data definition fails. Trying to -- existentially quantify K below seems to make -- sense, but also fails ... data Ainst b = I | J (Ainst b) | (B b) => K b instance (B b) => A (Ainst b) where a1 = I a2 = J a3 = K -- Reported line of the error data Binst = Val Int instance B Binst where b1 = Val ------------------------------- The error from ghci is as follows: Test5.hs:25:9: Couldn't match expected type `b' against inferred type `b1' `b' is a rigid type variable bound by the type signature for `a3' at Test5.hs:7:13 `b1' is a rigid type variable bound by the instance declaration at Test5.hs:16:12 Expected type: b -> Ainst b1 Inferred type: b1 -> Ainst b1 In the expression: K In the definition of `a3': a3 = K Failed, modules loaded: none. Thanks in advance for any help. Apologies if what I am doing is odd or the answer is obvious, I'm still very new to Haskell. Darryn.

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Re: [Haskell-cafe] "shadowing" keywords like "otherwise"
by Janis Voigtlaender 28 Jun '09

28 Jun '09

Vasili I. Galchin wrote: > so Janis I have to look where the original author overrode "otherwise"? > If so, just great .. a ton of code. ;^( No, you don't have to look for this. It is in the code snippet you sent: swishParse :: String -> String -> SwishStateIO (Maybe RDFGraph) swishParse fnam inp = do { fmt <- gets $ format ; case fmt of N3 -> swishParseN3 fnam inp otherwise -> do { swishError ("Unsupported file format: "++(show fmt)) 4 ; return Nothing } } In the case expression you use "otherwise" as a variable. But you do not use that variable in that branch. Hence the warning: Swish/HaskellRDF/SwishCommands.hs:304:12: Warning: Defined but not used: `otherwise' This has nothing to do with someone else shadowing the definition of "otherwise". Ciao, Janis. -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:voigt@tcs.inf.tu-dresden.de

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bizarre syntax error
by Geoffrey Irving 27 Jun '09

27 Jun '09

I ran into a unfortunate syntax error just now. I figured I'd share it because it's the weirdest message I've ever gotten out of ghc. The broken code is case t of TyApply tv types -> do (tvl, cases) <- lookupDatatype prog tv let tenv = Map.fromList (zip tvl types) -- GHC POINTS HERE caseType (c,vl,e') | Just tl <- List.lookup c cases = if length vl == a then expr prog global (foldl (\e (v,t) -> Map.insert v t e) env (zip vl (map (subst tenv) tl))) e' else typeError ("arity mismatch in pattern: "++show (pretty c)++" expected "++show a++" argument"++(if a == 1 then "" else "s") ++" but got ["++concat (intersperse ", " (map (show . pretty) vl))++"]") where a = length tl -- THIS IS THE PROBLEM | otherwise = typeError ("datatype "++show (pretty tv)++" has no constructor "++show (pretty c)) defaultType Nothing = return [] defaultType (Just (v,e')) = expr prog global (Map.insert v t env) e' >>=. \t -> [t] join t1 t2 | Just t <- unifyS t1 t2 = return t | otherwise = typeError ("failed to unify types "++show (pretty t1)++" and "++show (pretty t2)++" from different case branches") caseResults <- mapM caseType pl defaultResults <- defaultType def foldM1 join (caseResults ++ defaultResults) _ -> typeError ("expected datatype, got "++show (pretty t)) ghc complains that "The last statement in a 'do' construct must be an expression" and points to the line declaring "tenv". The actual problem is that you can't put a "where" block in between two pattern guards. :) Geoffrey

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ANNOUNCE: Cal3D animation library
by Gregory D. Weber 27 Jun '09

27 Jun '09

The Cal3D for Haskell project provides a partial binding to the C++ Cal3D animation library. The project homepage is http://haskell.org/haskellwiki/Cal3d_animation There are three packages available on hackage: * cal3d-0.1 is a Haskell binding to the Cal3D library itself. * cal3d-opengl-0.1 adds a few functions for using Cal3D with OpenGL. * cal3d-examples-0.1 provides a simple example based on the Cally Demo. Cal3D is a C++ library for skeletal-based character animation. It is platform-independent and independent of any particular graphics API. In fact, it does not actually render the animation, but provides hooks that the graphics API (such as OpenGL) can use to do the actual drawing. (As far as I know, though, the only graphics API available for Haskell is OpenGL.) For more information about the (C++) Cal3D library, please see * Cal3D FAQ: http://cal3d.sourceforge.net/docs/api/html/cal3dfaq.html * Cal3D Homepage: http://home.gna.org/cal3d/ -- ___ ___ __ _ / _ \ / _ \| | | | Gregory D. Weber, Associate Professor / /_\// / | | | /\ | | Indiana University East / /_\\/ /__| | |/ \| | http://mypage.iu.edu/~gdweber/ \____/\_____/\___/\__/ Tel. (765) 973-8420; FAX (765) 973-8550

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Trivial pivoting for the DSP lu decomposition
by Fernan Bolando 27 Jun '09

27 Jun '09

Hi all I created the beginnings of a simple circuit simulator using haskell. It can be downloaded from. http://plan9.bell-labs.com/sources/contrib/fernan/escomma.tar.bz2 It uses a modified version of the haskell DSP library matrix. I extended it with a simple pivoting Its not very haskelly, but it works with matrices that didn't work with the unmodified version. thanks for those that answered my list processing questions. I used that code to create a permutation vector for the pivoting. I am now looking into non-linear matrix solver. does anybody have a reference code for that? fernan -- http://www.fernski.com

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Rigid type variable error
by Darryn 27 Jun '09

27 Jun '09

From: Darryn <djreid(a)aapt.net.au> To: beginners(a)haskell.org Subject: Rigid type variables match error Date: Sat, 27 Jun 2009 10:18:13 +0930 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. Thanks in advance for any help anyone can provide.

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