- Haskell-Cafe - Haskell.org

Data representation, maybe reflection, laziness
by Mark Carroll 04 Nov '03

04 Nov '03

People have been talking about transmitting Haskell values on the GHC users' list, and I wanted to ask some more general stuff, partly out of mild ignorance. Ralf Hinze and Simon Peyton-Jones wrote an interesting paper on generic programming and derivable type classes. It looked like maybe programmers would be able to write their own "deriving xml" stuff and whatever, which looked great because, if there's not already one out there, I'd love to derive some read/show analogue automatically for data in some encoding that's very efficient to write and parse (i.e. not XML (-:). I was also wondering how the ability to write "deriving" stuff related to what one might think of as being reflection - that is, for example, could a definition of deepSeq be derived that automatically knows how to recurse into and traverse data structures and make sure everything's evaluated? This leads me to the point that little of the code I write needs laziness, but many of my unpleasant performance surprises come from laziness. I do find it hard to figure out where to put strictness annotations to actually make things work - for instance, I think it's laziness things that are causing my uses of Control.Exception.evaluate to actually work more by trial and error. No doubt it'll all grow clearer in time. Maybe I need "laziness annotations" instead. (-: Still, I was wondering what current thinking and directions were with regard to any of the above! -- Mark

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Doubt regarding Types
by Karthik Kumar 04 Nov '03

04 Nov '03

Hi All, I was trying to write go program in haskell - My code fragment looks like this - <---- > getBoardSize :: IO Bool -- TODO : What could be the type of getBoardSize getBoardSize = do c <- getLine validateBoardSize ( atoi c ) -- Convert a string to an integer. -- This works perfectly fine. atoi :: [Char] -> Int atoi (h : []) = if isDigit h then digitToInt h else 0 atoi (h : t) = if isDigit h then digitToInt h * ( 10 ^ length t) + atoi t else 0 -- validateBoardSize -- To validate the board size validateBoardSize :: Int -> Bool validateBoardSize d = (d == 9 || d == 13 || d == 19 ) <----> ERROR "test1.hs":21 - Type error in final generator *** Term : validateBoardSize (atoi c) *** Type : Bool *** Does not match : IO a Actually i guess there is something wrong with the type mentioned for getBoardSize function , but am clueless abt this. Thanks in advance for your help in this regard. Cheers Karthik __________________________________ Do you Yahoo!? Exclusive Video Premiere - Britney Spears http://launch.yahoo.com/promos/britneyspears/

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Help ? :)
by Patty Fong 04 Nov '03

04 Nov '03

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RE: fixed point
by Dean Herington 01 Nov '03

01 Nov '03

At 11:15 AM -0500 2003/10/31, Harris, Andrew wrote: > Hi - > I am trying to work out how the following function using "fix" is > evaluated. I am hoping someone could look at my step-by-step breakdown > of how I think evaluation works and see if I'm correct. My main > question is how the order of operation (fixity?) is understood in going > from step [3] to [4], if this is indeed how the evaluation would take > place. > > Any help is appreciated, > -andrew > > Here's the Haskell function: > > --- > fix f = f (fix f) > > wierdFunc2 x y z = if y - z > z then x (y-z) z > else y - z > > myRemainder = fix wierdFunc2 > --- > > Here's my "evaluation": You're pretty close. I've noted a couple of differences below. > myRemainder 12 5 -- [1] > = > fix wierdFunc2 12 5 -- [2] by substitution > = > wierdFunc2 fix wierdFunc2 12 5 -- [3] apply "fix" wierdfunc2 (fix wierdFunc2) 12 5 > = > wierdFunc2 myRemainder 12 5 -- [4] by substitution (?) No, equations "rewrite" only "from left to right". Skip the step above and the step below. > = > myRemainder 7 5 -- [5] apply "wierdFunc2" > = > fix wierdFunc2 7 5 -- [6] by substitution > = > wierdFunc2 fix wierdFunc2 7 5 -- [7] apply "fix" Similarly, the above needs (fix wierdFunc2) to be parenthesized. > = > wierdFunc2 myRemainder 7 5 -- [8] by substitution (?) Skip the above step. > = > 2 -- [9] apply "wierdFunc2" Dean

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newbie question regarding type declarations
by gangadhar npk 01 Nov '03

01 Nov '03

hi, I am a newbie to Haskell. I am reading the tutorial by Hal Daume III. Regarding the function types which are an extension of lambda calculus, I was wondering if at all it is possible to write a type that would return two values as the output - something like a square function which would take a pair and return a pair. I tried this, but there are errors square :: (Num a, Num b) => (a ,b) square (x , y) = (x*x , y*y) How can I specify that the square functions output would be a pair ? thank you gangadhar

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Monad @ Microsoft
by Josef Svenningsson 01 Nov '03

01 Nov '03

OK, this one is just too good not to post about it. Microsoft has developed a now command line interface. Guess what they call it? Read it here: http://slashdot.org/articles/03/10/31/1346201.shtml?tid=185&tid=190&tid=201 Well, it seems Simoj PJ is doing a good job introducting functional programming at MS although his methods are subtle in the extreme :-) /Josef

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RE: fixed point
by Harris, Andrew 01 Nov '03

01 Nov '03

Hi - I am trying to work out how the following function using "fix" is evaluated. I am hoping someone could look at my step-by-step breakdown of how I think evaluation works and see if I'm correct. My main question is how the order of operation (fixity?) is understood in going from step [3] to [4], if this is indeed how the evaluation would take place. Any help is appreciated, -andrew Here's the Haskell function: --- fix f = f (fix f) wierdFunc2 x y z = if y - z > z then x (y-z) z else y - z myRemainder = fix wierdFunc2 --- Here's my "evaluation": myRemainder 12 5 -- [1] = fix wierdFunc2 12 5 -- [2] by substitution = wierdFunc2 fix wierdFunc2 12 5 -- [3] apply "fix" = wierdFunc2 myRemainder 12 5 -- [4] by substitution (?) = myRemainder 7 5 -- [5] apply "wierdFunc2" = fix wierdFunc2 7 5 -- [6] by substitution = wierdFunc2 fix wierdFunc2 7 5 -- [7] apply "fix" = wierdFunc2 myRemainder 7 5 -- [8] by substitution (?) = 2 -- [9] apply "wierdFunc2"

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newbie question regarding type declarations
by gangadhar npk 31 Oct '03

31 Oct '03

hi, I am a newbie to Haskell. I am reading the tutorial by Hal Daume III. Regarding the function types which are an extension of lambda calculus, I was wondering if at all it is possible to write a type that would return two values as the output - something like a square function which would take a pair and return a pair. I tried this, but there are errors square :: (Num a, Num b) => (a ,b) square (x , y) = (x*x , y*y) How can I specify that the square functions output would be a pair ? thank you gangadhar

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Heap profiling in GHC broken?
by Markus.Schnell＠infineon.com 30 Oct '03

30 Oct '03

Hi. When I tried to profile my program in search for space leaks, it core dumped on me (actually on Win2000/Cygwin). I'm using ghc 6.0.1. Does anybody know what could be the problem? (To be more precise: any +RTS -h_ options make problems, +RTS -p works fine. I get a message meaning the equivalent of "statement in '0x0058db43' points to memory in '0xfffffffc'. Could not read.") Cheers, Markus -- Markus Schnell

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Haskell Asignement not working
by Rachel Reid 30 Oct '03

30 Oct '03

My code is --1)Define a function bLine to give a line of horizontal #s import Pictures sb, sw :: Picture bLine :: Int -> Picture bLine n ) |n==0 = ) |n>0 = `sideBySide` sb (bLine (n-1)) and when running Hugs, it says that there is a <unexpected symbol "#"> in line 9 which is bLine n ). There was a hash sign previously, but I have reloaded Hugs since then. can you help Thanks Rachel

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