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RE: automaticly create the ana, cata, hylomorphisms
by Marta Isabel Oliveira 07 Nov '03

07 Nov '03

Thanks and sorry. Beside new in Haskell i'm also new in this archive lists. So, i was not aware i had sent with sucess the mail to the first list. lesina > -----Original Message----- > From: Keith Wansbrough [SMTP:Keith.Wansbrough@cl.cam.ac.uk] > Sent: Friday, November 07, 2003 3:00 PM > To: Marta Isabel Oliveira > Cc: haskell(a)haskell.org; haskell-cafe(a)haskell.org > Subject: Re: automaticly create the ana, cata, hylomorphisms > > > Hello programmers, > > > > I have a problem to solve and don't have a clue where to start. > > This is a homework question. Please read > > http://www.haskell.org/hawiki/HomeworkHelp > > for some helpful hints. > > By the way, you shouldn't post the same message to multiple lists - you should choose the right one (haskell-cafe in this case), or if you must, cross-post one message to both. This means people who subscribe to both won't see your message twice. > > --KW 8-) > -- > Keith Wansbrough <kw217(a)cl.cam.ac.uk> > http://www.cl.cam.ac.uk/users/kw217/ > University of Cambridge Computer Laboratory. >

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Re: automaticly create the ana, cata, hylomorphisms
by Keith Wansbrough 07 Nov '03

07 Nov '03

> Hello programmers, > > I have a problem to solve and don't have a clue where to start. This is a homework question. Please read http://www.haskell.org/hawiki/HomeworkHelp for some helpful hints. By the way, you shouldn't post the same message to multiple lists - you should choose the right one (haskell-cafe in this case), or if you must, cross-post one message to both. This means people who subscribe to both won't see your message twice. --KW 8-) -- Keith Wansbrough <kw217(a)cl.cam.ac.uk> http://www.cl.cam.ac.uk/users/kw217/ University of Cambridge Computer Laboratory.

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Haskell pre-processor
by Marta Isabel Oliveira 07 Nov '03

07 Nov '03

Hello programmers, I have a problem to solve and don't have a clue where to start. I need to do a program in haskell that creates automaticly the ana, cata and hylomorphisms to arbitraty types. It's sort of a pre-processer. How can i start? I have the HaPar.hs that has uses Haskell parser to extract the type declarations from a file containing a Haskell module. I also have the TDec1.hs that has some dataTypes defined(TDec1, CDec1, Type) and some Show instances. Can anyone help me? I'm totally new to Haskell :(

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Another fold question
by Patty Fong 07 Nov '03

07 Nov '03

Having struglled over this for the better part of a day and only becoming more frustrated the more i try to understand it, i once again seek help :) I understand how basic folds work, i.e foldr replaces (:) with some parameter and [] by another i.e foldr (+) 0 [1,2,3] becomes 1+(2+(3+0)) I also understand how to write my own fold function. What i don't understand quite is how to use them. given this data type and this fold function i wrote: data Music = Note Pitch Octave Duration | Silence Duration | PlayerPar Music Music | PlayerSeq Music Music | Tempo (Ratio Int) Music data Pitch = Cf | C | Cs type Octave = Int type Duration = Ratio Int foldMusic :: (Pitch -> Octave -> Duration -> a) -> (Duration -> a) -> (a -> a -> a) -> (a -> a -> a) -> (Ratio Int -> a -> a) -> Music -> a foldMusic n _ _ _ _ (Note pitch octive duration) = n pitch octive duration foldMusic _ s _ _ _ (Silence duration) = s duration foldMusic n s p1 p2 t (PlayerPar partOne partTwo) = p1 (foldMusic n s p1 p2 t partOne)(foldMusic n s p1 p2 t partTwo) foldMusic n s p1 p2 t (PlayerPar partA partB) = p2 (foldMusic n s p1 p2 t partA)(foldMusic n s p1 p2 t partB) foldMusic n s p1 p2 t (Tempo rate part) = t rate (foldMusic n s p1 p2 t part) I understand that when i use the foldMusic function i need to pass it 5 parameters. given the type signiature, why can i pass (+) as a parameter for p1 but not for n, what determines what can be passed as a parameter, because they all have the return type a?? I attempted to create a function that utilises the foldMusic function that counts the number of notes: count_notes :: Music -> Integer count_notes = foldMusic (\_-> \_ -> \_ -> 1) (\_ -> 0) (+) (+) (\_ -> \_ -> 0) it appears to work, i think. Yet i'm still not certain of how it does so. This confuses me, Is there anyway to represent other fold functions in a tree like representation as foldr (+) 0 would appear as such? + 1 \ + 2 \ + 3 \ 0 Regards, confused Patrick _________________________________________________________________ Hot chart ringtones and polyphonics. Go to http://ninemsn.com.au/mobilemania/default.asp

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Another Newbie question :)
by Patty Fong 06 Nov '03

06 Nov '03

I seem to understand the basic idea behind the fold function, i think but do not seem to be able to write them myself. My understanding is that it works by replacing every constructor with a fold ? Could anyone please point me in the direction of a suitable resource of attempt to explain this to me :) Cheers, Paty _________________________________________________________________ Hot chart ringtones and polyphonics. Go to http://ninemsn.com.au/mobilemania/default.asp

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Trying to understand tuples and lists
by Karthik Kumar 05 Nov '03

05 Nov '03

Hi, I have a list of tuples ( characters and numbers ) . say, [('a', 1), ('b', 2)] I want to extract 'a' and 1 separately . (x,y) <- head [('a', 1), ('b', 2)] This doesnt work but my idea is to get x contain the value 'a' and y contain the value 1 as the first tuple in the list. Which is the right way of doing this. Thanks for your help. Cheers Karthik. __________________________________ Do you Yahoo!? Protect your identity with Yahoo! Mail AddressGuard http://antispam.yahoo.com/whatsnewfree

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Type errors in Haskell programming languages - Plz help
by Karthik Kumar 05 Nov '03

05 Nov '03

Hi, I was trying this Haskell program and was getting the following type error. The code is as follows - <-- Code starts --> entry :: [Char] -> [(Char,Int)] entry list = do t <- getGroups list mergeGroups t getGroups :: [Char] -> [(Char,Int)] mergeGroups :: [(Char,Int)] -> [(Char,Int)] <-- Code Ends --> The function getGroups and mergeGroups compile perfectly fine. The error i am getting is in the function entry . To paste the message, " *** Expression : mergeGroups t *** Term : t *** Type : (Char,Int) *** Does not match : [(Char,Int)] " Thanks in advance for helping me in this regard. Cheers Karthik. __________________________________ Do you Yahoo!? Protect your identity with Yahoo! Mail AddressGuard http://antispam.yahoo.com/whatsnewfree

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RE: Type tree traversals [Re: Modeling multiple inheritance]
by Simon Peyton-Jones 05 Nov '03

05 Nov '03

| We really should change GHC rather than keep trying to work around stuff | like this. GHC will be my light reading for winter break. Maybe so. For the benefit of those of us who have not followed the details of your work, could you summarise, as precisely as possible, just what language extension you propose, and how it would be useful? A kind of free-standing summary, not assuming the reader has already read the earlier messages. Simon

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Re: Data types basics
by Patty Fong 05 Nov '03

05 Nov '03

Another prompt reply! thanks Hal :) I think I understand this correctly now For my previous problem which was: <prolog> ::= (<assertion> ".")* <assertion> :: = <structure> | <rule> <rule> ::= <structure> ":-" <structure>("," <structure>)* <structure> ::= <name> [( <term> (, <term>)* )] <term> ::= <number> | <variable> | <structure> <variable> ::= <name> Does this appear to be correct: type Prolog = [Assertion] data Assertion = StrucAssert Structure | RuleAssert Rule data Rule = Rule Structure Structure [Structure] data Structure = Structure Name Term [Term] data Term = NumTerm Number | VarTerm Variable | StrucTerm Structure type Variable = Name type Name = String type Number = Int >From: Hal Daume III To: Patty Fong CC: Haskell Cafe , Hal Daume Subject: >Re: Data types basics Date: Tue, 4 Nov 2003 22:10:03 -0800 (PST) > >Hi again, > >On Wed, 5 Nov 2003, Patty Fong wrote: > > > Hi to anyone reading this. i'm still strugling a bit with data type > >declarations. > > The was i understand it is that if i delcare a new data >type: > > data myType = myType a | b | c > >This isn't entirely correct. The names of types have to begin with capital >letters, as do constructors (more later). So you would need this to be: > >data MyType = MyType A | B | C > >where A is an existing type. > >This type now has three constructors: > >MyType :: A -> MyType B :: MyType C :: MyType > >It's perhaps a bit easier to understand when the names are different. > >When we say: > >data Foo = Bar Int | Baz String > >this means that a "Foo" is of one of two forms (the "|" can be read as >disjunction). A value of type Foo is either of the form "Bar x" for some x >which is an Int or "Baz y" for some y which is a String. > >"Bar" and "Baz" are called "constructors" because they take arguments and >"construct" a Foo. So, in this case, > >Bar :: Int -> Foo Baz :: String -> Foo > >are the two constructors. > >Of course, you have have any number of constructors and each can have any >number of arguments. > >data Foo = Bar | Baz Int | Bazaa String Bool > >Now there are three constructors: > >Bar :: Foo Baz :: Int -> Foo Bazaa :: String -> Bool -> Foo > >they can be recursive: > >data Foo = Bar | Baz Int Foo > >Bar :: Foo Baz :: Int -> Foo -> Foo > >and can have "type variables", for instance: > >data Foo a = Bar | Baz a > >here, something of type "Foo a" is either of the form Bar or of the form >Baz x for some x which is of type a. This has constructors: > >Bar :: Foo a Baz :: a -> Foo a > >I hope this sheds some light on the issue... > > >-- >Hal Daume III | hdaume(a)isi.edu "Arrest this man, he talks in maths." | >www.isi.edu/~hdaume > _________________________________________________________________ Hot chart ringtones and polyphonics. Go to http://ninemsn.com.au/mobilemania/default.asp

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Data types basics
by Patty Fong 05 Nov '03

05 Nov '03

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