November 2011 - Beginners

Re: [Haskell-beginners] Beginners Digest, Vol 41, Issue 33
by Google Wave 24 Nov '11

24 Nov '11

2011/11/24 <beginners-request(a)haskell.org> > Send Beginners mailing list submissions to > beginners(a)haskell.org > > To subscribe or unsubscribe via the World Wide Web, visit > http://www.haskell.org/mailman/listinfo/beginners > or, via email, send a message with subject or body 'help' to > beginners-request(a)haskell.org > > You can reach the person managing the list at > beginners-owner(a)haskell.org > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of Beginners digest..." > > > Today's Topics: > > 1. Basic "data" question (Costello, Roger L.) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Thu, 24 Nov 2011 10:43:58 +0000 > From: "Costello, Roger L." <costello(a)mitre.org> > Subject: [Haskell-beginners] Basic "data" question > To: "beginners(a)haskell.org" <beginners(a)haskell.org> > Message-ID: > <B5FEE00B53CF054AA8439027E8FE1775181911E9(a)IMCMBX04.MITRE.ORG> > Content-Type: text/plain; charset="us-ascii" > > Hi Folks, > > I am trying to define a data type for this: > > A Lambda term is one of these: > - Variable > - Application of a Lambda term to a Lambda term > - Abstraction of a Variable in a Lambda term > > Here's an example of a Lambda term: > > term = Abstraction (Variable X) (Application (Variable X) (Variable Y)) > > Is there a way to define Lambda term? > > Here's an attempt at defining it: > > data Var = V | W | X | Y | Z > > data Lambda = Variable Var | Application Lambda Lambda | Abstraction > (Variable Var) Lambda > > But that yields the following error: > > test.hs:5:71: Not in scope: type constructor or class 'Variable' > > Any suggestions would be much appreciated. > > /Roger > > > > ------------------------------ > > _______________________________________________ > Beginners mailing list > Beginners(a)haskell.org > http://www.haskell.org/mailman/listinfo/beginners > > > End of Beginners Digest, Vol 41, Issue 33 > ***************************************** > value constructor followed by types. And (Variable Var) belong to Lambda type. Could define like this: data Lambda var= Variable var | Application (Lambda var) (Lambda var) var be any types

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Re: [Haskell-beginners] Basic "data" question
by Brent Yorgey 24 Nov '11

24 Nov '11

On Thu, Nov 24, 2011 at 03:01:49PM +0000, Costello, Roger L. wrote: > Hi Brent, > > Thanks for the feedback. > > > data Var = V | W | X | Y | Z > > > > data Lambda = Variable Var | Application Lambda Lambda | Abstraction Var Lambda > > Unfortunately, that's not quite correct. The first argument of > Abstraction must be a Lambda term, specifically Variable Var. No, that isn't possible (and it also doesn't make sense). It isn't possible because there is no way to express "this field may only contain values built with such-and-such constructor".* It doesn't make sense because why must the first argument of Abstraction be a lambda term? You yourself said "Abstraction of a Variable in a Lambda term" -- it is not "abstraction of a lambda term in a lambda term", but abstraction of a *variable*. Variables have type Var. Hence, Abstraction Var Lambda (I can also use an argument from authority: I have coded many variants of the lambda-calculus in Haskell, and I can tell you that this is the way it is done. =) -Brent * Actually, it is sometimes possible to encode this using GADTs. But that wouldn't help you here.

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Basic "data" question
by Costello, Roger L. 24 Nov '11

24 Nov '11

Hi Folks, I am trying to define a data type for this: A Lambda term is one of these: - Variable - Application of a Lambda term to a Lambda term - Abstraction of a Variable in a Lambda term Here's an example of a Lambda term: term = Abstraction (Variable X) (Application (Variable X) (Variable Y)) Is there a way to define Lambda term? Here's an attempt at defining it: data Var = V | W | X | Y | Z data Lambda = Variable Var | Application Lambda Lambda | Abstraction (Variable Var) Lambda But that yields the following error: test.hs:5:71: Not in scope: type constructor or class 'Variable' Any suggestions would be much appreciated. /Roger

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help opengl object
by Song Zhang 23 Nov '11

23 Nov '11

Hi in Graphics.UI.GLUT.Objects there are some pre-defined object such as cube or cone I tried to draw objects in haskell but get black screen who can give me some examples to draw these objects and for drawing Cylinder in the documentation what is (*freeglut only) *means. do I need glut32.dll to render these object in windows? thanks -- the University of Nottingham, CS school the representative of SSFC CS school Staff of SIFEUNNC IT department I am Ice which is sleeping water. Sent via BlackBerry© by BerryMail

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Re: [Haskell-beginners] data design for a questionnaire (retitled + update)
by Alia 23 Nov '11

23 Nov '11

As a followup, I just received a simpler and more elegant solution due to Lorenzo Bolla which involves wrapping the question type in another type and achieves the objective nicely. I think this is much cleaner than using existential types as per my prior posting. <version4.hs> module Main where import Text.Show.Functions -- type converters str = id int s = read s :: Int float s = read s :: Double data QuestionType = Open | Test | Choice deriving (Show, Eq) data Question a = Question { questionName :: String , questionText :: String , questionType :: QuestionType , answerFunc :: (String -> a) , correctAnswer :: Maybe a , options :: Maybe [(String, a)] } deriving (Show) data Question' = QuestionS (Question String) | QuestionI (Question Int) deriving (Show) data QuestionSet a = QuestionSet { qsetTitle :: String , qsetQuestions :: [Question'] } deriving (Show) data Questionnaire a = Questionnaire { questionnaireTitle :: String , questionnaireQuestionSets :: [QuestionSet a] } deriving (Show) q1 = Question { questionName = "q1" , questionText = "What is our name?" , questionType = Open , answerFunc = id , correctAnswer = Nothing , options = Nothing } q2 = Question { questionName = "q2" , questionText = "What is 1+1?" , questionType = Test , answerFunc = int , correctAnswer = Just 2 , options = Nothing } q3 = Question { questionName = "q2" , questionText = "What is 2+1?" , questionType = Choice , answerFunc = int , correctAnswer = Just 3 , options = Just [("a", 2), ("b", 3), ("c", 4)] } qset = QuestionSet { qsetTitle = "simple questions" , qsetQuestions = [QuestionS q1, QuestionI q2, QuestionI q3] } questionnaire = Questionnaire { questionnaireTitle = "a questionnaire" , questionnaireQuestionSets = [qset] } </version4.hs> Alia

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List comprehension question
by Lee Short 23 Nov '11

23 Nov '11

So hlint suggests that I convert the line checked = if ival==input then [("checked","checked")] else [] to checked = [("checked","checked") | ival==input] I'm not really sure why this works. Is there some implicit list being filtered by ival==input? What's going on? thanks Lee

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Precedence of Infix Operators in Do Syntax
by Avery Robinson 23 Nov '11

23 Nov '11

Hello, I was reading this block of code from http://jasani.org/2008/02/18/command-line-haskell-and-error-handling-exampl… main = do m <- hGetContents stdin nums <- mapM readM . lines $ m print (sum nums) `catch` (\e -> hPutStrLn stderr ("couldn't sum lines: " ++ show e)) readM :: (Monad m, Read a) => String -> m a readM s | [x] <- parse = return x | otherwise = fail $ "Failed to parse \"" ++ s ++ "\" as a number." where parse = [x | (x,_) <- reads s] I don't understand how line 5 works. I thought that the do notation there would be the same as: main = hGetContents stdin >>= \m -> mapM readM . lines $ m >>= \nums -> print (sum nums) >> `catch` (\e -> hPutStrLn stderr ("couldn't sum lines: " ++ show e)) But I don't understand how that makes any sense with that infix `catch`. It looks like catch doesn't even have a first argument. Please enlighten me. Thanks.

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Re: [Haskell-beginners] Beginners Digest, Vol 41, Issue 30
by Lee Short 22 Nov '11

22 Nov '11

> On Tue, 22 Nov 2011 14:46:37 -0800 > Lee Short <blackcat(a)pro-ns.net> wrote: > >> hlint gives me a parse error on a clause using DoIfThenElse, even >> if >> I have the language pragma. I don't see any hlint options to get >> around that, are there any? >> >> Is it considered good style to write code like this? >> >> if "" == results >> then return True >> else return False >> >> The obvious way rewrite below just seems clunky to me (though I can >> see how others might prefer it to the code above). >> >> return $ if "" == results >> then True >> else False > > You've just pressed one of my language-independent style hot > buttons. Why on earth are you using an if/then/else here? What's > wrong > with the straightforward: > > return "" == results > > The expression results in a boolean, and it's even the one you want > to > return. So why not return it? > That's not my actual code, I was just using it as a simplified example and didn't put a lot of thought into it. My actual code is if delete /= "" then if exists then deleteFromTable req t key else return () else if exists then updateTable req t key else insertTable req t I'd love to hear comments on how to make that more idiomatic, if there's a way.

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hlint and DoIfThenElse
by Lee Short 22 Nov '11

22 Nov '11

hlint gives me a parse error on a clause using DoIfThenElse, even if I have the language pragma. I don't see any hlint options to get around that, are there any? Is it considered good style to write code like this? if "" == results then return True else return False The obvious way rewrite below just seems clunky to me (though I can see how others might prefer it to the code above). return $ if "" == results then True else False thanks Lee

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questionnaire data design patterns
by Alia 22 Nov '11

22 Nov '11

Hi folks, As a way to get my head wrapped around haskell data design idioms, I'm translating a questionnaire data model (in python) to haskell, I would appreciate some advice and criticism on whether I'm going about this the right way. Here's the initial translation: <version1.hs> module Main where data QuestionType = Open | Test | Choice deriving (Show, Eq) data Question = Question { questionName :: String , questionText :: String , questionType :: QuestionType , answerType :: String , correctAnswer :: Maybe String , options :: Maybe [(String, String)] } deriving (Show, Eq) data QuestionSet = QuestionSet { qsetTitle :: String , qsetQuestions :: [Question] } deriving (Show, Eq) data Questionnaire = Questionnaire { questionnaireTitle :: String , questionnaireQuestionSets :: [QuestionSet] } deriving (Show, Eq) q1 = Question { questionName = "q1" , questionText = "What is our name?" , questionType = Open , answerType = "str" , correctAnswer = Nothing , options = Nothing } q2 = Question { questionName = "q2" , questionText = "What is 1+1?" , questionType = Test , answerType = "int" , correctAnswer = Just "2" , options = Nothing } q3 = Question { questionName = "q2" , questionText = "What is 2+1?" , questionType = Choice , answerType = "int" , correctAnswer = Just "3" , options = Just [("1", "2"), ("2", "3"), ("3", "4")] } qset = QuestionSet { qsetTitle = "simple questions" , qsetQuestions = [q1, q2, q3] } questionnaire = Questionnaire { questionnaireTitle = "a questionnaire" , questionnaireQuestionSets = [qset] } </version1.hs> In this first version, the Question record basically holds all possible fields for any kind of question, and fields which are not used are assigned Nothing thanks to the Maybe monad. The final type of the answer is specified as a string in the record for later conversion. Now I wanted to take into account a scenario where I don't need to specify the answerType by creating a polymorphic Question type which is specialized to take different answertypes. So I came up with this version: <version2.hs> module Main where import Text.Show.Functions -- type converters str = id int s = read s :: Int float s = read s :: Double data QuestionType = Open | Test | Choice deriving (Show, Eq) data Question a = Question { questionName :: String , questionText :: String , questionType :: QuestionType , answerFunc :: (String -> a) , correctAnswer :: Maybe a , options :: Maybe [(String, a)] } deriving (Show) data QuestionSet a = QuestionSet { qsetTitle :: String , qsetQuestions :: [Question a] } deriving (Show) data Questionnaire a = Questionnaire { questionnaireTitle :: String , questionnaireQuestionSets :: [QuestionSet a] } deriving (Show) q1 = Question { questionName = "q1" , questionText = "What is our name?" , questionType = Open , answerFunc = id , correctAnswer = Nothing , options = Nothing } q2 = Question { questionName = "q2" , questionText = "What is 1+1?" , questionType = Test , answerFunc = int , correctAnswer = Just 2 , options = Nothing } q3 = Question { questionName = "q2" , questionText = "What is 2+1?" , questionType = Choice , answerFunc = int , correctAnswer = Just 3 , options = Just [("a", 2), ("b", 3), ("c", 4)] } qset = QuestionSet { qsetTitle = "simple questions" , qsetQuestions = [q1, q2, q3] } questionnaire = Questionnaire { questionnaireTitle = "a questionnaire" , questionnaireQuestionSets = [qset] } </version2.hs> The problem is that now all the questionsets, and questionnaires expect the same specialized type of question which is obviously not what I want and I get this error: Couldn't match expected type `[Char]' with actual type `Int' Expected type: Question String Actual type: Question Int In the expression: q2 In the `qsetQuestions' field of a record So what am I doing wrong here? Should I just revert back to version 1 which I sense is not the haskell way. Many thanks for any help or advice. Alia

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