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Re: Type equivalency 2
by Ashley Yakeley 06 Jun '02

06 Jun '02

At 2002-06-06 00:40, Cagdas Ozgenc wrote: >From the previous discussion it has been brought to my attention that there >is no much difference between > >a -> b > >and > >data F a b = Blank a b -- "-> probably has a blank value constructor" Well F and (->) may both be type-constructors of kind (* -> * -> *), but I think there the similarity ends. You may consider an F as holding an 'a' and a 'b', but a function holds a 'b' for every 'a'. It's quite different, and F does not implement functions. If you're wondering about the value constructor for a function, ask yourself, what would you pass to it? You could do this, though: newtype F' a b = MkF' (a -> b) ...or this: type F'' = (->) I think you'll find (->) is pretty much atomic. It's tempting to say a function is constructed from a piece of code, but for a lazy language pretty much any value might be represented by a piece of code... -- Ashley Yakeley, Seattle WA

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Re: type equivalency
by Jon Cast 06 Jun '02

06 Jun '02

Andrew J Bromage <andrew(a)bromage.org> wrote: > G'day all. > On Wed, Jun 05, 2002 at 08:20:03PM -0500, Jon Cast wrote: > > I think you're confused about what the type declarations mean. > > When you say > > > sqrt :: Float -> Float > > you're promising to operate over /all/ Floats. > That would be true of Haskell functions were constrained to be total > functions. They are not. Sqrt takes values of type Float, but it > just happens to be a partial function over that type. That depends on what you mean by ``partial function''. Technically, of course, a function of type (a -> b) is a subset of (a x b) satisfying certain constraints. All Haskell functions f satisfy the property (forall x::a. exists y::b. (x, y) <- f), so you can always get a value of f at any x. In other words, sqrt can take any x and compute with it; it just delivers _|_ at some of them. My point was, (sqrt x) is legal Haskell (and has a value) for all (x :: Float). > > Unfortunately, Haskell doesn't allow {x :: Float | x >= 0} as a > > type, nor does it provide a positive-only floating point type. > One general rule of strongly-typed programming is: A program is type > correct if it is accepted by my favourite type checker. A corollary > is that what you call a type, I reserve the right to call a > precondition. If I accepted that, I would be un-defining crucial terms. That would destroy the potential for discussion here, no? > Cheers, > Andrew Bromage Jon Cast

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Re: type equivalency
by Jon Cast 06 Jun '02

06 Jun '02

Cagdas Ozgenc <co19(a)cornell.edu> wrote: > For example all functions with Int -> Int are type equivalent, > because structural equivalency is used. Most of the time the > functions are not compatible because they have different pre/post > conditions, even though they seem to have the same input output > types. I rather make my functions an instance of a type-class to > tell the compiler that they can be used interchangably. For example > sqrt :: Float -> Float > sqrt a = /* pre : a >= 0 */ > negate :: Float -> Float > /* no precondition */ > are different functions. You cannot just plug one in place of > another (you can in Haskell because of structural type equivalency). I think you're confused about what the type declarations mean. When you say > sqrt :: Float -> Float you're promising to operate over /all/ Floats. If you want to restrict that to only certain Floats, you would need a different ``type'': > sqrt :: {x :: Float | x >= 0} -> Float or introduce your own type isomorphic to this. Unfortunately, Haskell doesn't allow {x :: Float | x >= 0} as a type, nor does it provide a positive-only floating point type. I think your problem is related to the concepts of ``different'' and ``plug in''. All the Haskell compiler performs is type checking; unfortunately, type checking is purely structural; it doesn't enquire whether the resulting program is correct (or even meaningful). You can tell Haskell isn't concerned with whether your programs are meaningful, because Haskell allows the following program: > main = main This is obviously completely meaningless. Now, then, obviously sqrt and negate are different in the sense of ``unequal''. If you don't know, equality between Haskell functions is defined (ommitting some complications) like such: > f = g <=> forall x.f x = g x Clearly negate /= sqrt! However, taking your notion of ``not different'', which is apparantly > f = g <=> domain f = domain g note that both > sqrt x = x ** (1/2) and > qrrt x = x ** (1/4) have the same domain. Are they equivalent? It (obviously) depends on the context. Consider the function dist: > dist :: (Double, Double) -> (Double, Double) -> Double > dist (x1, x2) (y1, y2) = sqrt ((x1 - y1) ** 2 + (x2 - y2) ** 2) You can't use qrrt in dist and expect it to be correct any more than you an use negate. On the other hand there may be terms which don't care which among qrrt and sqrt you give them but don't want negate. So, equivalence is a tricky problem---and one you can't solve without reading the programmer's mind. That's why Haskell doesn't address it, being content with type checking. > Also a function working over (Int,Int) will do so even if the > numbers are totally irrelevant to that function. They maybe the > number of (apples,oranges) or number of (books,authors). You're right that the type constructor (,) doesn't carry a lot of information; but I think you misunderstand why: given a type declaration > data FiddleDeDee a b = MkFiddleDeDee a b FiddleDeDee is ``structurally equivalent'' to (,), and also carries about as much information. Consider: ``What's the French for fiddle-de-dee?'' ``Fiddle-de-dee's not English,'' Alice replied gravely. ``Who ever said it was?'' said the Red Queen. Lewis Carroll, ``Through the Looking Glass'' > However, > data D a b = MkD a b > data E a b = MkE a b > (D Int Int) and (E Int Int) are trated as different because of name > equivalency testing. > Basically there are 2 type constructors -> ( , ) with structural > equivalency, which is odd. Someone just dumped some idiosyncracies > of lamda calculus into the language. I assume from your example that by ``structural equivalence'' you mean the relationship between D and E. But they also share that relationship with (,)---what exactly is the difference? Functions over (D Int Int) work even if the caller thinks its (D #apples #oranges) and the function thinks its (D #books #authors). Again, the problem is with Int, not D (or (,)). > Thanks again.

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RE: Counting occurrences question
by Andy Fugard 05 Jun '02

05 Jun '02

>===== Original Message From "xoo" <ixoo21(a)hotmail.com> ===== >hi.. i was just wondering if some body could give a simple equation for the following situation.other than recursion plz.. > >occurrences :: Eq a => a -> [a] -> [a] >--occurrences xs ys returns the number of times that xs occurs in ys You may find it easier if you make occurrences :: Eq a => a -> [a] -> Integer since it would seem it is to return a number, and not another list! Also I would guess the function will have a form something like occurrences x xs = foldr (countOp x) 0 xs where countOp :: Eq a => a -> a -> Integer -> Integer ... I'm not sure if I should go any further, in case this is a homework question... Andy -- [ Andy Fugard ] [ +44 (0)7901 603075 ]

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Bilgisayar Egitiminde Bir Numara
by Sertifikam.Com 05 Jun '02

05 Jun '02

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Bilgisayar Egitiminde Bir Numara
by Sertifikam.Com 05 Jun '02

05 Jun '02

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by targetemailextractor＠btamail.net.cn 03 Jun '02

03 Jun '02

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HTTP Implementation
by Warrick Gray 02 Jun '02

02 Jun '02

I've been tinkering with a client-side HTTP/1.1 implementation for Haskell. My ambition is to write a small Haskell program for checking my Hotmail account. In the interests of finding all bugs and giving something for nothing I've made the code available: http://homepages.paradise.net.nz/warrickg/haskell/http Significantly missing is any hint of SSL and request pipelining. Suggestions on improvements are welcome. Warrick. _________________________________________________________________ Send and receive Hotmail on your mobile device: http://mobile.msn.com

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RE: Naming conventions for compiler options [Was: layout rule infelicity]
by Jay Allen 31 May '02

31 May '02

-Wmissing-prototypes, actually (http://www.esat.kuleuven.ac.be/~gcc/) But good guess. ;) -J- -----Original Message----- From: Ashley Yakeley [mailto:ashley@semantic.org] Sent: Thu 5/30/2002 5:26 PM To: Haskell Cafe List Cc: Subject: Re: Naming conventions for compiler options [Was: layout rule infelicity] At 2002-05-30 16:51, oleg(a)pobox.com wrote: >-Wmissing-protons Compiles BASIC? -- Ashley Yakeley, Seattle WA _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe(a)haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe

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Re: Naming conventions for compiler options [Was: layout rule infelicity]
by Ashley Yakeley 31 May '02

31 May '02

At 2002-05-30 16:51, oleg(a)pobox.com wrote: >-Wmissing-protons Compiles BASIC? -- Ashley Yakeley, Seattle WA

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Naming conventions for compiler options [Was: layout rule infelicity]
by oleg＠pobox.com 31 May '02

31 May '02

[redirected to haskell-cafe] Jón Fairbairn wrote: > 1. Why "-f" anyway? It took me ages to work out what > "-fallow-overlapping-instances" meant -- I wondered how > "fallow" could apply to overlapping instances. I believe the authors of GHC followed the naming conventions of GCC, which can be gleaned from GCC's info page. -warning-related options typically start with -W -machine-independent flags that control optimization, code-generation or language dialects start with -f. Most flags have positive and negative forms. -many pre-processor-related options start with -i -hardware-dependent options start with -m, e.g., -mcpu=i686 -mno-fancy-math-387 -malign-int -mno-power [RS/6000-related option] -myellowknife [On embedded PowerPC systems, assume that the startup module is called `crt0.o' and the standard C libraries are `libyk.a' and `libc.a'] Of possible interest is a file gcc-2.95.2/gcc/future.options from GCC's source code. The file lists a few suggested options. The file is an e-mail message from Noah Friedman to Richard Stallman, Jim Blandy and a few other people. Some of the suggested options are: -Waggravate-return -Wcast-spell -Wcaste-align [cf. the existing GCC option -Wcast-align] -Win [I guess warn about the Windows system] -Wmissing-protons -Wredundant-repetitions -antsy -fbungee-jump -fexpensive-operations [cf. -fexpensive-optimizations] -fextra-strength [must be a negative form of -fstrength-reduce] -fkeep-programmers-inline [cf. the existing option -fkeep-inline-functions] -fjesus-saves [cf. the existing option -fcaller-saves] -fno-peeping-toms [cf. the existing option -fno-peephole] -fruit-roll-ups [cf. the existing option -funroll-loops] -fshort-enough -mno-dialogue -vomit-frame-pointer

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Re: layout rule infelicity
by Frank Atanassow 30 May '02

30 May '02

[redirected to haskell-cafe] Ashley Yakeley wrote (on 30-05-02 03:18 -0700): > At 2002-05-30 02:54, Lennart Augustsson wrote: > > >If you look at C (& offspring), it's not the {;} that makes the code > >readable, it's the indentation that does. So why not acknowledge that? > > In C, the indentation is an important visual clue, but there are many > different indentation styles. I think there are not so many different _indentation_ styles. The differences seem mostly to revolve around where to put _the braces_. And one might argue that the reason for that is precisely that it's so arbitrary, since the indentation is the main visual clue to structure anyway. Personally, I always try to factor out arbitrariness from my programs. I think layout does the same thing for syntax. > If you're used to braces, complicated Haskell expressions with layout > look confusing, since it's not immediately clear which indentation style > the layout rules are trying to enforce. If you're used to C, then layout and indentation will be the least of your difficulties when you start using Haskell... Anyway, I have the feeling that, for every person on this list who complains about layout being unintuitive, there are 10 people who would say the opposite. Shall we take a poll? -- Frank Atanassow, Information & Computing Sciences, Utrecht University Padualaan 14, PO Box 80.089, 3508 TB Utrecht, Netherlands Tel +31 (030) 253-3261 Fax +31 (030) 251-379

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Space problem: reading a list of records from a binary database
by Eray Ozkural 29 May '02

29 May '02

Hi there, While trying to read a list of records from a binary database I think I've encountered a space leak but I'm not proficient enough to detect the cause of the error. The culprit is the readSessionDB function in SessionDB module which I attach with other relevant code. What I do in Session.readSession is using FFI to read a fixed size record from an input stream. That far, okay. (I have no idea if I'm doing it right, though) The problem is, combined with my approach in Session.readSessionDB, it results in huge amounts of space allocation, about 1MB, whereas the raw data would consume about 50K. It would be a [Session] with about 3000 elements, each of which contains 3 Ints and 6 Floats. I wonder if the following code in SessionDB is the proper way to read a list of records from a file. I get the feeling that the (s:rest) in the last return statement causes my problems. readSessionList :: Handle -> IO [Session] readSessionList h = do session <- readSession h case session of Nothing -> return [] Just s -> do rest <- readSessionList h return (s:rest) Comments appreciated, -- Eray Ozkural (exa) <erayo(a)cs.bilkent.edu.tr> Comp. Sci. Dept., Bilkent University, Ankara www: http://www.cs.bilkent.edu.tr/~erayo Malfunction: http://mp3.com/ariza GPG public key fingerprint: 360C 852F 88B0 A745 F31B EA0F 7C07 AE16 874D 539C

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RE: toEnum with international chars
by Simon Marlow 28 May '02

28 May '02

> Oh, my bad, I think I gave a wrong list there. 66 is of > course 'B', I was a > little sleepless at the time of writing. Now having enjoyed a > few hours of > sleep I can express more eloquently :) > > It seems to be the fourth character that is the problem, but > then again that's > my mistake. Thank you. > > Still it attracts my attention that I cannot type þ in ghci > while I can type > it in bash. I think I can't type any of those international > chars in ghc. > Maybe some bug fix is in order. Perhaps you could give an example? Bash and GHCi both use the Readline library, and as far as I can tell you can type exactly the same characters to both. Cheers, Simon

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toEnum with international chars
by Eray Ozkural 28 May '02

28 May '02

I have a code that reads a string from a list of CChar's > shortName = map (toEnum.fromIntegral) shortname When I have a turkish character like Ý the interpreter will: StockList> a <- readStockInfo h [66,105,108,105,-2,105,109,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32,32] StockList> a *** Exception: Prelude.chr: bad argument The problem is the 66 in the beginning of the list. I'm not giving the entire code, it's not needed... What should I be using to work with such strings? Thanks, -- Eray Ozkural (exa) <erayo(a)cs.bilkent.edu.tr> Comp. Sci. Dept., Bilkent University, Ankara www: http://www.cs.bilkent.edu.tr/~erayo Malfunction: http://mp3.com/ariza GPG public key fingerprint: 360C 852F 88B0 A745 F31B EA0F 7C07 AE16 874D 539C

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Critique required
by Andy Fugard 27 May '02

27 May '02

Hi all, I'm a Haskell newbie. Recently I threw together an answer to Exercise 3.2 on p47 of Hudak, which is attached below. Essentially all it does is draw a star, then on each point of the star draw another star, and so on, recursively. Would it be possible for someone to comment briefly on my answer. For example is there anything in there that is immediately annoying to an experienced Haskell programmer? How are programs usually commented? (I come from an imperative background, where functions are often longer -- there it makes sense to write pre- and post-conditions for each function, but I'm not sure if that's relevant for Haskell.) Any ideas for making it more efficient? So, essentially I'm asking someone to rip me to shreds :) I don't yet have any technical queries, but have just reached the section on "fold", so I expect it won't be long.... Cheers, Andy ---------------------------------------------------------- import SOEGraphics intToFloat :: Int -> Float intToFloat n = fromInteger (toInteger n) starOfDavidList :: Point -> Int -> [Point] starOfDavidList (x,y) side = [ ( x, y - halfHeight ), ( x - halfWidth, y + halfHeight ), ( x + halfWidth, y + halfHeight ), ( x, y + halfHeight + offset ), ( x - halfWidth, y - halfHeight + offset ), ( x + halfWidth, y - halfHeight + offset ) ] where s = intToFloat side height = sqrt (s^2 - (s^2 / 4)) halfHeight = round (height/2) halfWidth = round (s/2) offset = round (1/3 * height) drawStar :: Window -> Color -> Point -> Int -> IO () drawStar w c pos side = do drawInWindow w equiTriUp drawInWindow w equiTriDown where equiTriUp = withColor c (polygon (take 3 list)) equiTriDown = withColor c (polygon (drop 3 list)) list = starOfDavidList pos side snowFlake :: Window -> [Color] -> Int -> Int -> Point -> IO () snowFlake w colors side itr pos | length colors < itr = error "Not enough colours in list" | itr > 0 = do drawStar w (head colors) pos side sequence_ (map funct ptList) | otherwise = return () where funct = snowFlake w (tail colors) newWidth (itr - 1) ptList = starOfDavidList pos side newWidth = round (1/3 * intToFloat(side)) main = runGraphics( do w <- openWindow "A Snowflake Fractal" (700,650) snowFlake w [Cyan, Blue, Magenta, White, Blue] 400 5 (300,250) k <- getKey w closeWindow w ) -- [ Andy Fugard ] [ +44 (0)7901 603075 ]

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Constructing Cases
by Carl McTague 27 May '02

27 May '02

Hi there, is there a simple way to carry out the following type of conversion? Suppose I have a list (finite) of (value,image) pairs such as: list = [(0,1),(1,0)] >From this I want to generate a function f 0 = 1 f 1 = 0 Is there a way to do this? Note, I don't want to define a function that searches through the list each time it is invoked, I want to generate the function once and have it be as fast as the pattern-matcher can make it. So, I'm looking for a function g :: [(a,b)] -> a -> b Is this simple to do? Thanks, Carl

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Segmentation fault
by ck15 23 May '02

23 May '02

Hi, I am getting a segmentation fault , when I execute a code in hugs. (only some times) What are the conditions that give rise to a segmentation fault , when executing a Haskell code in hugs? I would appreciate very much if some could spare some time to help me. Thanks Chitra

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�𾴵��Ͽͻ��Ǻ�!
by web 22 May '02

22 May '02

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RE: Re: Fixed point of n-mutually recursive functors
by gonflores＠guay.com 22 May '02

22 May '02

> I don't think this is possible in plain Haskell 98. > It is possible in Generic Haskell, see: > http://www.generic-haskell.org/ > [...] > Generic Haskell's type system is sufficient. > > Is it possible using dependent types? > > I would expect so. I have been looking to Generic Haskell and it may be possible to define a generic map function like: mapN :: (a1 -> b1) -> ... -> (aN -> bN) -> f a1 ... aN -> f b1 ... bN However, I still don't know how could I define something like: > data FixN1 f1 ... fN = InN1 (f1 (FixN1 f1 ... fN) ... (FixNN f1 ... fN)) > ... > data FixNN f1 ... fN = InNN (fN (FixN1 f1 ... fN) ... (FixNN f1 ... fN)) I think that I would need dependent types, because I want to define a number of data types that depend on a given natural...any suggestions? Best regards, Gonzalo Flores

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convering Fds to Handles
by Abraham Egnor 22 May '02

22 May '02

Is there a way to convert an Fd variable to a Handle variable? Specifically, I'd like to use "pipe" from the Posix library to create a pipe and then pass one end of it to runProcess as its stdin. Abe

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Fixed point of n-mutually recursive functors
by gonflores＠guay.com 22 May '02

22 May '02

I am able to define the fixed point of a functor as > data Fix f = In (f (Fix f)) where I can also define the type class > class Functor1 f > where map1 :: (a -> b) -> f a -> f b I can also define the fixed-point of 2 mutually recursive bifunctors as: > data Fix21 f g = In21 (f (Fix21 f g) (Fix22 f g)) > data Fix22 f g = In22 (g (Fix21 f g) (Fix22 f g)) where bifunctors can be captured in the type class > class Functor2 f > where map2 :: (a1 -> b1) -> (a2 -> b2) -> f a1 a2 -> f b1 b2 Now, I would like to generalize those definitions to n-functors. Is it possible? If it isn't, what kind of type system would I need? Is it possible using dependent types? I would also appreciate any points to literature using n-mutually recursive functors. Best regards, Gonzalo Flores

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derivative (S f g) ?
by Conal Elliott 21 May '02

21 May '02

This isn't really a Haskell question, but I'm hoping a fellow Haskeller might have some helpful pointers. Has anyone seen a generalization of the chain rule for derivatives that applies to applications of the S combinator? The conventional chain rule applies to the more restricted composition combinator: D (f . g) = (D f . g) * D g Where D is the differentiation higher-order function, "*" is multiply lifted pointwise to functions (\ a b x -> a x * b x), and "." is function composition. Thanks, - Conal

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(no subject)
by Andy Fugard 21 May '02

21 May '02

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Re: SUGGESTION: haskell-announce mailing list
by Jens Petersen 21 May '02

21 May '02

"Simon Marlow" <simonmar(a)microsoft.com> writes: > I'm happy to set up and moderate > haskell-announce(a)haskell.org if there's a general > consensus that this would be a useful thing. I think there are enough announcements on the haskell list, to warrant having a separate announce list. It would make filtering announcements to a separate folder much easier too. > As to whether we should merge haskell and haskell-café - > personally I wasn't in favour of the split at the time, > but I recall that support was roughly 50/50 in favour. > Unless there's an overwhelming majority in favour of a > merge I suggest we leave things as they are. I don't have a strong opinion on this either way, however I agree with some of the earlier comments that it only really makes sense to have two different lists if their agenda are more clearly distinct. (IMHO "cafe" is too vague.) Jens

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19 May '02

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Re: Q: Resolving ambiguous type variable
by tty＠SAFe-mail.net 17 May '02

17 May '02

Would it be correct to conclude from your reasoning that the following would be properly typed by the compiler ? first_h :: (Show a) => [a] -> Result a first_h [] = Ex "No list" first_h (x:xs) = Value x since we are explicitly stating first_h can only take a list containing elements deriving Show. My expectations would be that this would be unambiguous, unfortunately ghc-5.02.3 still complains with the reported error. Regards Tee -------- Original Message -------- From: hdaume(a)ISI.EDU To: tty(a)SAFe-mail.net Cc: haskell-cafe(a)haskell.org Subject: Re: Q: Resolving ambiguous type variable Date: Fri, 17 May 2002 10:24:29 -0700 (PDT) > Erm, I think I said something stupid (or at least implied it). > > main = getArgs >>= print . first_h > > *will* work, because it knows that the result of a getArgs is a list of > Strings. > > The problem you originally had was that when you say: > > first_h [] > > The "[]" could refer to a list of any type of element. Since not every > element type is an instance of Show, it doesn't know that it can apply > show to the list. > > By saying, for instance, (first_h ([] :: [Char])) you're saying "okay, > this is an empty list, but it's a list of Characters and you know that a > character can be shown." If you backtrack what the type checker is doing, > it says: > > print (first_h []) > > okay, the argument to print must be an instance of Show (print has type > Show a => a -> IO ()). therefore, > > first_h [] > > must be an instance of Show. well, what's the type of first_h []? Well, > the type of first_h is [a] -> Result a. is "Result a" an instance of > show? well, let's see...you said data Result a = ... deriving > (Show). this basically means you get an instance declaration: > > instance Show a => Show (Result a) where > ...some stuf... > > so, we want to know if Result a is an instance of show. Well, it is > whenever a is an instance of show. so, if the elements of your list are > an instance of show, then (first_h []) is an instance of Show, as we > require. > > however, since you just say "[]" it knows this is of type [a], but it > *doesn't* know what a is. therefore, a could either be or not be an > instance of show. however, if you explicitly specify what a is and that > explicit type is an instance of show (like Char in my example), then > everything is fine. > > -- > Hal Daume III > > "Computer science is no more about computers | hdaume(a)isi.edu > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > On Fri, 17 May 2002, Hal Daume III wrote: > > > In short, you cannot. > > > > What if your main were: > > > > main = getArgs >>= print . first_h > > > > The compiler doesn't know the difference and so it needs a type. > > > > Simple fix: > > > > > main = print (first_h ([] :: [Char])) > > > > > > -- > > Hal Daume III > > > > "Computer science is no more about computers | hdaume(a)isi.edu > > than astronomy is about telescopes." -Dijkstra | www.isi.edu/~hdaume > > > > On Fri, 17 May 2002 tty(a)SAFe-mail.net wrote: > > > > > Hello, > > > I am writing a function which returns an exception/error string should an unexpected parameter is passed in. Borrowing from the Maybe module I have the following: > > > ------------------------------------------------------------ > > > data Result a = Ex String | Value a deriving Show > > > > > > -- Testing the Result a type > > > first_h :: [a] -> Result a > > > first_h [] = Ex "No list" > > > first_h (x:xs) = Value x > > > > > > -- Trying out the first_h > > > main :: IO() > > > main = print (first_h []) > > > ------------------------------------------------------------ > > > > > > Which the compiler complains: > > > > > > Ambiguous type variable(s) `a' in the constraint `Show a' > > > arising from use of `print' at atype.hs:8 > > > In the definition of `main': print (first_h []) > > > > > > This is understandable since it is unable to unify the empty list with a concrete list of type 'a' i.e. there are an infinite types which would match. My question is how can I indicate to the compiler that it does not matter which type the empty list is since the return result is of a concreate type. > > > > > > Thanks > > > > > > Tee > > > _______________________________________________ > > > Haskell-Cafe mailing list > > > Haskell-Cafe(a)haskell.org > > > http://www.haskell.org/mailman/listinfo/haskell-cafe > > > > > > > _______________________________________________ > > Haskell-Cafe mailing list > > Haskell-Cafe(a)haskell.org > > http://www.haskell.org/mailman/listinfo/haskell-cafe > >

2 1

Q: Resolving ambiguous type variable
by tty＠SAFe-mail.net 17 May '02

17 May '02

Hello, I am writing a function which returns an exception/error string should an unexpected parameter is passed in. Borrowing from the Maybe module I have the following: ------------------------------------------------------------ data Result a = Ex String | Value a deriving Show -- Testing the Result a type first_h :: [a] -> Result a first_h [] = Ex "No list" first_h (x:xs) = Value x -- Trying out the first_h main :: IO() main = print (first_h []) ------------------------------------------------------------ Which the compiler complains: Ambiguous type variable(s) `a' in the constraint `Show a' arising from use of `print' at atype.hs:8 In the definition of `main': print (first_h []) This is understandable since it is unable to unify the empty list with a concrete list of type 'a' i.e. there are an infinite types which would match. My question is how can I indicate to the compiler that it does not matter which type the empty list is since the return result is of a concreate type. Thanks Tee

2 2

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by tony 17 May '02

17 May '02

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1 0

Simple Question
by Jerry, JiJie 16 May '02

16 May '02

Hi, enlightenment much appreciated for my newbie question -- to achieve: append' [ [1, 2], [3, 4], [5] ] 6 -> [ [1, 2], [3, 4], [5, 6] ] append' [ ['1', '2'], ['3'] ] '4' -> [ ['1', '2'], ['3', '4'] ] append' [ [True], [True] ] False -> [ [True], [True, False] ] so I (naively) write: 18) append' :: [[a]] -> a -> [[a]] 19) append' [] y = [[y]] 20) append' x:xs y = [(init x:xs)] ++ [(tail xs)++[y]] and ghc(5.02.3) says: $ ghc p3a.hs p3a.hs:20: Parse error in pattern thanx in advance! -- Regards, Jerry

2 3

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by raeline 16 May '02

16 May '02

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1 0

��ȫ��ͼ�
by raeline 15 May '02

15 May '02

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RE: Replacing the Prelude
by Simon Peyton-Jones 14 May '02

14 May '02

Ashley writes | > I was hoping to do something similar for 'do' notation by redefining | > (>>), (>>=) etc., but unfortunately GHC is quite insistent | that 'do' notation quite specifically refers to GHC.Base.Monad Dylan replies | I'm surprised that ghc uses the fromInteger and fromRational | that are in scope; I thought there was general agreement that | it should use the Prelude.from{Integer,Rational} in scope. Ashley is referring to a GHC extension. Normally, GHC uses Prelude.fromInteger etc, regardless of what is in scope. But if you say -fno-implicit-prelude, GHC will instead use whatver fromInteger is in scope. (This is documented in the manual.) Ashley's question, as I understand is whether something similar could be done for monads. Answer: I think so, without much bother. I'm beavering away on a Haskell workshop paper at the moment, but ping me in a fortnight to do it. Simon

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Re: Replacing the Prelude
by Dylan Thurston 14 May '02

14 May '02

On Sun, May 12, 2002 at 09:31:38PM -0700, Ashley Yakeley wrote: > I have recently been experimenting writing code that replaces large > chunks of the Prelude, compiling with -fno-implicit-prelude. I notice > that I can happily redefine numeric literals simply by creating functions > called 'fromInteger' and 'fromRational': GHC will use whatever is in > scope for those names. > > I was hoping to do something similar for 'do' notation by redefining > (>>), (>>=) etc., but unfortunately GHC is quite insistent that 'do' > notation quite specifically refers to GHC.Base.Monad (i.e. Prelude.Monad, > as the Report seems to require). I don't suppose there's any way of > fooling it, is there? I was rather hoping 'do' notation would work like a > macro in rewriting its block, and not worry about types at all. > > I accept that this might be a slightly bizarre request. There are a > number of things I don't like about the way the Prelude.Monad class and > 'do' notation are set up, and it would be nice to be able to experiment > with alternatives. A while ago, there were extensive discussions about replacing the Prelude on this list. (Search for "Prelude shenanigans".) I started to write up a design document for how to enable replacing the Prelude. This boiled down to taking most of the "syntactic sugar" defined in the report seriously, ignoring the types (as you say). I'm surprised that ghc uses the fromInteger and fromRational that are in scope; I thought there was general agreement that it should use the Prelude.from{Integer,Rational} in scope. As I recall, there were several relevant bits of syntactic sugar: - Numeric types, including 'fromInteger', 'fromRational', and 'negate'. This all works fine, except that the defaulting mechanism is completely broken, causing a number of headaches. - Monads. The translation given in the report is clean, and it seems like it can be used without problems. - Bools. There was a slight problem here: the expansion of 'if ... then ... else ...' uses a case construct referring to the constructors of the Bool type, which prevents any useful redefinition of Bool. I would propose using a new function, 'Prelude.ifThenElse', if there is one in scope. - Enumerations. Clean syntactic sugar. - List comprehensions. The report gives one translation, but I think I might prefer a translation into monads. - Lists and tuples more generally. At some point the translations start getting too hairy; I think I decided that lists and tuples were too deeply intertwined into the language to change cleanly. I'll dig up my old notes and write more, and then maybe write a complete design document and get someone to implement it. --Dylan Thurston

1 0

composeList
by Emre Tezel 13 May '02

13 May '02

Hi all, I recently bought Simon Thompson's Haskell book. I have been doing the exercises while I read on. There are couple questions that I can not solve. Any help would be greatly appreciated. I got stuck on this question in Chapter 10. Define a function composeList which composes a list of functions into a single function. What is the type of composeList? I naively tried the following but the hugs compiler complained about the inferred type not being generic enough. composeList :: [(a -> b)] -> (c -> b) composeList [] = id composeList (x:xs) = x . (composeList xs) or composeList :: [(a -> b)] -> (c -> b) composeList (x:xs) = x . (composeList xs) ERROR "Chapter10.hs":6 - Inferred type is not general enough *** Expression : composeList *** Expected type : [a -> b] -> c -> b *** Inferred type : [a -> a] -> b -> a Anybody can solve this? Emre

2 2

Re: waiting for your reply
by Bob Johnson 12 May '02

12 May '02

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1 0

preprocessing printf/regex strings (like ocaml)
by Oliver George 12 May '02

12 May '02

Hi Haskellers, I'm interested to know why a string translating preprocessor doesn't exist for haskell. It seems to me that it would alleviate printf and regex like problems in an convenient, elegant and type-safe manner. An example of this I came across recently was the ocaml printf statement: # Printf.printf "roar";; roar- : unit = () # Printf.printf "numbers %d %d" 11 23;; numbers 11 23- : unit = () # Printf.printf "a number %d" "word";; This expression has type string but is here used with type int You can see logically how this might work in haskell, compile time evaluation would translate the string into the appropriate native expressions before the statements type is evaluated. I see a similar possibility for regular expression matching, wouldn't it be nice in a haskell script to write: perl like things... msg' = replace "s/love/lust/" msg or, nice regex stuff... main = case match "^(\d+)" of Nothing -> 0 Just (i) -> i the python string notation (str % tuple) would fit really well too... putStrLn "hello %s, you got %d right" % ("oliver", 5) Am I the only one who sees this as being a really valuable extension to haskell? or does it exist and i've just never noticed? cheers, Oliver.

2 1

Wishing you a wonderful day
by Bob Johnson 12 May '02

12 May '02

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1 0

examples of functional objects
by Pixel 11 May '02

11 May '02

It seems to be hard to have something both abstract and concrete. http://merd.net/pixel/language-study/various/non-mutable-objects-and-subtyp… The first version (file haskell.listing) is quite simple, but the last one (file haskell3.listing) gets pretty hairy: ---------------------------------------- class Show a => Point a where translate :: a -> Double -> a my_length :: a -> Double add :: a -> a -> a pack :: a -> Packed_point unpack :: a -> Points pack = Packed_point data Packed_point = forall a. Point a => Packed_point a data Points = TPoint2d Point2d | TPoint3d Point3d data Point2d = Point2d(Double, Double) deriving Show data Point3d = Point3d(Double, Double, Double) deriving Show ---------------------------------------- is there a simpler solution? thanks! PS: please, don't give me the simple non-OO version, it's already there (haskell3bis.listing), and I give an explaination why i think it's not enough: non-OO Haskell and OCaml versions are the simplest, but do not allow precise enough typing: no distinction between Point2d and Point3d is done at compile time. So the only way to have a function working only on Point3d's is to raise an error when that function is called with a Point2d (see get_z) (of course, compared to dynamically typed languages versions, it's already much better)

1 0

ObjectIO install
by Andre W B Furtado 10 May '02

10 May '02

The INSTALL file that comes with the objectIO files says that one must use the following commands to register the package: $ ghc-pkg -r objectio $ ghc-pkg -a objectio -f pckg.cfg The second command is not recognized as a valid one, and the ghc-pkg help appears showing the correct syntax. I needed to use $ ghc-pkg -a < pckg.cfg in order to get the package correctly registered. Am I missing something or is the INSTALL file wrong? -- Andre

2 1

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by sales137＠163.com 09 May '02

09 May '02

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RE: Using "type"
by Simon Peyton-Jones 09 May '02

09 May '02

GHC currently takes the following (non-Haskell 98) attitude to type synonyms: first expand type synonyms only then check for partial applications of type synonyms then perform type checking So in his case type for f expands to: f :: m () -> m () test :: Int -> Int You can see how the type for test comes out: Thing (Const Int) = Const Int () = Int Now GHC tries to unify m () with Int and fails, as it should. The error message is phrased in terms of the original un-expanded types, which is usually a good thing but on this occasion gives a rather odd looking error message. GHC shouldn't really do this lazy checking unless you say -fglasgow-exts Simon | -----Original Message----- | From: Samuel E. Moelius III [mailto:usmoeliu@mcs.drexel.edu] | Sent: 04 May 2002 20:45 | To: haskell-cafe(a)haskell.org | Subject: Using "type" | | | I apologize if this question has been asked before... | | It seems to me the following code should be legal: | | type Thing m = m () | | type Const a b = a | | f :: Thing m -> Thing m | f x = x | | test :: Thing (Const Int) -> Thing (Const Int) | test = f | | However, GHC gives the following error: | | Couldn't match `Thing m' against `Thing (Const Int)' | Expected type: Thing m | Inferred type: Thing (Const Int) | Expected type: Thing (Const Int) -> Thing (Const Int) | Inferred type: Thing m -> Thing m | | But if you change the definition of Const | | newtype Const a b = MakeConst a | | the code compiles fine. | | Is this the desired behavior? | | Is it the case that when you have quantification of a type | variable of | kind * -> * (in this case, m), that variable may only become | bound to a | type constructor? | | Sam Moelius | | _______________________________________________ | Haskell-Cafe mailing list | Haskell-Cafe(a)haskell.org | http://www.haskell.org/mailman/listinfo/haskell-cafe |

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RE: Deforestration harmful?
by Simon Marlow 08 May '02

08 May '02

> I will be a bit more precise. Let f and g have type: > > g :: Seed -> [Tree] > f :: Tree -> [Expr] > > where Tree1 is a tip-valued binary tree, and Tree2 is also a > binary tree but the forks are arithmetic operators: > > data Tree = Tip Int | Bin Tree1 Tree1 > data Expr = Val Int | Add Tree2 Tree2 | Sub Tree2 Tree2 | Mul .... These don't make sense. Perhaps for Tree1 you mean Tree, and Tree2 you mean Expr? > The function g is defined like: > > g seed = [ z | (s1, s2) <- split seed, > x <- g s1, y <- g s2, z <- Bin x y ] There's a type error in that definition: 'Bin x y' is not a list. > The function f converts a Tree to many Exprs by replacing > Bin with one of Add, Sub, Mul or Div: > > f (Tip i) = [Val i] > f (Bin x y) = [op x' y' | x' <- f x, y' <- f y, op <- join x y] > > The program before fusion looks like: > > concat . map f . g > > while the fused program simply: > > fg seed = [ z | (s1, s2) <- split seed, > x <- fg s1, y <- fg s2, z <- join x y ] > join x y = [op x y | op <- [Add, Sub, ... ]] > > And the surprise was the fused program ran slower than the > naive one before fusion. > > The time profiling showed that, in one run, the naive > program itself took 2.00 sec to finish, with 9% extra GC time. The > fused program itself, on the other hand, took only 1.84 sec but > spent an extra 67% of time on GC (the heap size is the default > 64M). > > Considering their heap profiles, both of them look like several > steep, slanted triangles standing next to each other. The highest peak > in the heap profile of the naive program reaches only 50K of memory, > while that for the fused program reaches as much as 1500K. In > the former case, the heap is mostly occupied by chunks created > by g. In the latter case, the heap is mostly occupied by join. > > > Furthermore, deforestation itself does not reduce runtime very much. > > Indeed. Looks like it helps to speed up the mutator a bit but > end up spending more time on GC, due to increased heap demand. I think this is a misleading generalisation - in most cases deforestation doesn't affect the heap residency behaviour of the program. As Olaf said, deforestation tends to result in small constant factor speedups, but can be particularly beneficial if it enables further optimisations to take place. In your case it looks like the transformation you made by hand had the side-effect of introducing a space leak, but it's quite hard to tell from the code you've shown us. (space leaks are particularly hard to identify by eye, using the profiling tools is the only way to really understand what's happening). > Monday, May 6, 2002, 12:02 PM, Simon Marlow wrote： > > It is entirely plausible that as you increase the heap > size the mutator > > time increases: this is because you get fewer cache hits > with a larger > > heap. GHC starts off with a small heap size so as to try > to make the > > most of the cache. However, on most examples we've seen > the benefits of > > reducing garbage collections by increasing the heap size tend to > > outweigh the benefits from staying in the cache. > > Did you try heap profiling to discover what structure was > filling up > > the heap? > > Thanks for the explanation. :) > > More info. about the heap was given above. By the way, could there be > situations where increasing the heap size also increases the GC time? Yes, again because of increased cache misses. Cheers, Simon

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Re: Proper scaling of randoms
by Dylan Thurston 07 May '02

07 May '02

(Redirected to haskell-cafe.) On Mon, May 06, 2002 at 04:49:55PM -0700, oleg(a)pobox.com wrote: > Problem: given an integer n within [0..maxn], design a scaling > function sc(n) that returns an integer within [s..t], t>=s. > The function sc(n) must be 'monotonic': > 0<=a < b ==> sc(a) <= sc(b) > and it must map the ends of the interval: > sc(0) -> s and sc(maxn) -> t. Just an aside (which Oleg surely knows): for actual random number generation, you often don't care about the monotonicity, and only care about uniform generation. In this case, there is a very simple algorithm: work modulo (s-t+1). scm(n) = (n `rem` (s-t+1)) + s Warning: some, broken, random number generators do not behave well when used like this. Also, although this is as uniform as possible, there is a systematic skew towards the lower end of the range [s..t]. --Dylan Thurston

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administrative question
by Matthias 07 May '02

07 May '02

dear haskell-o-tronics, >From time to time i get emails from haskell-cafe, with the subject "RE: <some topic>" but since i am only subscribe since some days i dont have the original message and thus would be habby if anybody could bounce me the list-mails from the last year? Matthias

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Re: Bug in the scaling of randoms ...
by Ronny Wichers Schreur 07 May '02

07 May '02

Dimitre Novatchev writes (to the Haskell mailing list): > In his excellent book Simon Thompson defines scaling of the elements of > a sequence of random numbers from the interval [0, 65535] to an > interval [s,t] in the following way (top of page 368): > > scaleSequence :: Int -> Int -> [Int] -> [Int] > > scaleSequence s t > > = map scale > > where > > scale n = n `div` denom + s > > range = t - s + 1 > > denom = modulus `div` range > > where modulus = 65536 > However, the following expression: > > e4000 = (scaleSequence 1 966 ([65231] ++ [1..965]))!!0 > evaluates to 974 -- clearly outside of the specified interval. The function fails in this case because the range doesn't divide the modulus. In the book (I checked the Miranda version), there's a remark after the function that "[The] range of numbers 0 to modulus-1 is split into range blocks, each of the same length." When the range doesn't divide the modulus, it's not possible to get a uniform distribution if you scale each value individually. I assume this is acceptable, otherwise you'll have to discard or combine values from the original sequence. Oleg answers: >We aim to find a function sc(n) defined over integers 0 <= n <= M so >that > sc(0) -> s (given integral number) > sc(M) -> t (given integral number), t>=s >and > 0<=a < b ==> sc(a) <= sc(b) > >Such a function sc(n) is given by the following expression: > sc(n) = (n*(t-s)) `div` M + s (M = modulus - 1) I think a more natural function would be sc(n) = (n * (t-s+1)) `div` modulus + s This satifies the conditions (assuming range = t-s+1 <= modulus), and distributes the values more evenly, in particular when the range divides the modulus. Cheers, Ronny Wichers Schreur

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Re: Using "type"
by Ashley Yakeley 05 May '02

05 May '02

At 2002-05-04 12:45, Samuel E. Moelius III wrote: > type Thing m = m () > > type Const a b = a > > f :: Thing m -> Thing m > f x = x > > test :: Thing (Const Int) -> Thing (Const Int) This is wrong. Type synonyms must be fully applied, whereas you have Const applied only to one argument, not two. 'Const' isn't really a type-constructor, it's more like a macro. Is it something in the air? Spring is here, and everyone wants type-lambda... -- Ashley Yakeley, Seattle WA

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Using "type"
by Samuel E. Moelius III 04 May '02

04 May '02

I apologize if this question has been asked before... It seems to me the following code should be legal: type Thing m = m () type Const a b = a f :: Thing m -> Thing m f x = x test :: Thing (Const Int) -> Thing (Const Int) test = f However, GHC gives the following error: Couldn't match `Thing m' against `Thing (Const Int)' Expected type: Thing m Inferred type: Thing (Const Int) Expected type: Thing (Const Int) -> Thing (Const Int) Inferred type: Thing m -> Thing m But if you change the definition of Const newtype Const a b = MakeConst a the code compiles fine. Is this the desired behavior? Is it the case that when you have quantification of a type variable of kind * -> * (in this case, m), that variable may only become bound to a type constructor? Sam Moelius

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Re: tricky instance declarations
by Ashley Yakeley 03 May '02

03 May '02

At 2002-05-03 03:53, Warrick Gray wrote: >instance Monad (IO (Either ConnError a)) where You do of course mean instance Monad (/\a => IO (Either ConnError a)) where ... ...but you just can't do type-lambdas in Haskell. -- Ashley Yakeley, Seattle WA

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User-Defined Types
by Matthias 01 May '02

01 May '02

Hello, I am a student of computer science at the university of Aachen in Germany and presently do a course about functional programming in haskell. I just wanted to introduce myself, because it is likely to happen that my posting apear frequently in the next couple of month. So my first question is about User-Defined Types. Stating the Gentle Introduction to Haskell 98 (2.2) , i already know that: "Type Constructors and date Constructors are in seperte namespaces. [example]" But according to the chapter (2.2.1) "Recursive Types" i see the polymorphic definition of a tree is: data Tree a = Leaf a | Branch (Tree a) (Tree a) where (Tree a) is obviously the data-constructor out of "the other namespace". I guess that the () is warping Tree a from the "data-constructor-namespace" into the "type-constructor-namespace". Since the gentle-intro implicitly uses (), which i am not used to - i would be happy about some hints about (). That might not be to difficult, i hope. Also i would not regret some hints about "->" which is used in function-type-definitions. I would prefer writing plus :: a,a -> a instead of the ->-variant in order to stress the distinction between arguments and returned type. What is meant that -> is "associating to the right"? Is there a name for '->'? Sincerely Matthias Pfeifer

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