June 2002 - Haskell-Cafe

typeclass versioning
by Cagdas Ozgenc 12 Jun '02

12 Jun '02

Greetings. What happens if a type is made an instance of a typeclass in two different modules with different implementations? In static linking scenarios it will be caught by the compiler. I don't know whether we can do dynamic loading in Haskell, but it seems there will be inconsistencies in that case. Even in static linking when an object is serialized over the wire where a different implementation of a typeclass exists things will go wild. What's Haskell's approach to versioning? Thanks

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Generating the n! permutations in Haskell
by Shlomi Fish 10 Jun '02

10 Jun '02

Included below is a Haskell Program I wrote to generate the n! permutations of a set of n elements. There are comments that explain it throughout the program. Both of the algorithms that appear in the cut the knot site use assignment, so are not very suitable for Haskell. Regards, Shlomi Fish -- (gradual_transfer set empty_set) -- -- Gradually pops elements out of set and [1..5] pushes them into -- empty_set, -- and makes a list of both stacks in their intermediate phases. -- -- E.g: -- gradual_transfer [1 .. 5] [] = -- [([1,2,3,4,5],[]),([2,3,4,5],[1]),([3,4,5],[2,1]), -- ([4,5],[3,2,1]),([5],[4,3,2,1])] gradual_transfer :: [a] -> [a] -> [([a],[a])] -- I stop when the list contains a single element, not when it contains -- no elements at all. The reason for this is that gen_perms like it -- better -- this way, as it has no use of a zero element (a:as). gradual_transfer (a:[]) ps = [((a:[]),ps)] gradual_transfer (a:as) ps = ((a:as),ps):(gradual_transfer as (a:ps)) -- (dump ps as) is equivalent to (reverse ps) ++ as, only it should -- be much faster. dump :: [a] -> [a] -> [a] dump [] as = as dump (p:ps) as = dump ps (p:as) gen_perms :: [a] -> [[a]] gen_perms [] = [[]] gen_perms set = [ (a:rest) | (a:as,ps) <- (gradual_transfer set []), rest <- gen_perms(dump ps as) ] print_perms [] = return () print_perms (a:as) = do print a print_perms as main = print_perms (gen_perms [1 .. 8]) ---------------------------------------------------------------------- Shlomi Fish shlomif(a)vipe.technion.ac.il Home Page: http://t2.technion.ac.il/~shlomif/ Home E-mail: shlomif(a)iglu.org.il He who re-invents the wheel, understands much better how a wheel works.

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Re: HTTP.hs with slow execution
by Warrick Gray 09 Jun '02

09 Jun '02

[redirecting to haskell-cafe] Hi Jens, I think I have it. >I think I found an issue with the HTTP.hs library when >making a short-lived non-persistent connection to a server >from a slower machine (i486). I can also reproduce it by >running my program with strace or ltrace on a faster machine >(PIII). Last two meaningful recv call: >recvfrom(3, 0x500a3878, 1000, 0, 0x500a3c68) = 555 >getpeername(3, 0x500a3c8c, 0x500a3ca4, 0x080dfda0, 1) = 0 >recvfrom(3, 0x500a2008, 1000, 0, 0x500a23f8) = 0 >getpeername(3, 0x500a241c, 0x500a2434, 0x080dfda0, 1) = 0 After no data is received the sending part of the TCP stream is closed: >shutdown(3, 1, 0x0805b85f, 0x080e018c, 1) = 0 Being polite, the library ensures that all incoming data has been consumed, forgetting for a moment that the incoming stream has already closed: >recvfrom(3, 0x500a2440, 1000, 0, 0x500a2830) = 0 >getpeername(3, 0x500a2854, 0x500a286c, 0x080e018c, 1) = -1 The "recvfrom" seems fine, the "getpeername" call comes from the definition of Network.Socket.recvFrom, which features: flg <- sIsConnected sock -- For at least one implementation (WinSock 2), recvfrom() ignores -- filling in the sockaddr for connected TCP sockets. Cope with -- this by using getPeerName instead. sockaddr <- if flg then getPeerName sock else peekSockAddr ptr_addr ...Leading to the error in question: >strerror(107) = "Transport endpoint is >not connec"... The answer I guess is to alter all calls to 'recvFrom' to 'recv' calls (done), or replace socket reference with ConnClosed at the first sign of a closed socket. Unfortunately the second will have to wait, but the first is implemented and available in the latest version: http://homepages.paradise.net.nz/warrickg/haskell/http Warrick. PS - nice trace. _________________________________________________________________ Get your FREE download of MSN Explorer at http://explorer.msn.com/intl.asp.

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problems figuring out what the type system is telling me
by Chris Moline 08 Jun '02

08 Jun '02

hi. i am really stuck on this problem and i cant seem to figure it out and so i am posting this in the hopes someone will help me. i realize you prolly get this kind of question alot but the wiki and the papers i have been reading arent helping to clarify things. i am trying to write a little utility to make it easier to get rid of ports on my machine. the function that is giving me trouble is getDepends. it takes the name of a port and opens the appropriate +CONTENTS file and returns the dependencies it finds. here is the function and its helper. getDepends :: String -> [String] getDepends p = do handle <- openFile (portsDir ++ p) ReadMode fetchDepends handle fetchDepends :: Handle -> [String] fetchDepends handle = do l <- hGetLine handle e <- hIsEOF handle case (not e) of -- ifs keep giving indent errors so ill just use case True -> case (matchRegex (mkRegex "^@pkgdep") l) of Just [a] -> [drop 8 l] ++ (fetchDepends handle) _ -> fetchDepends handle False -> "" here is ghci's error messages. Compiling Pheobe ( Phoebe.hs, interpreted ) Phoebe.hs:19: Couldn't match `[]' against `IO' Expected type: [t] Inferred type: IO Handle In the application `openFile (portsDir ++ p) ReadMode' In a 'do' expression pattern binding: handle <- openFile (portsDir ++ p) ReadMode Phoebe.hs:24: Couldn't match `[]' against `IO' Expected type: [t] Inferred type: IO String In the application `hGetLine handle' In a 'do' expression pattern binding: l <- hGetLine handle Failed, modules loaded: none. could someone be so kind as to explain what the problem is? sincerly, chris moline

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type equivalency
by Cagdas Ozgenc 06 Jun '02

06 Jun '02

Greetings. I know 2 special type constructors(there might be other that I do not know yet) -> and ( , ) where structural type equivalency is enforced and we can also create new types with an algebric type constructor notation where name equivalency is enforced. What is the rationale? I mean why 2 special type constructors, but not 5, or 10 or N? Thanks for taking time.

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(no subject)
by xoo 06 Jun '02

06 Jun '02

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 thanks

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

06 Jun '02

At 2002-06-06 03:38, Cagdas Ozgenc wrote: >Then either -> is not a type constructor, (->) is a type constructor. >or the concept of type >constructors has to be divided into two : consuming type constructors, and >producing type constructors. Type constructors can do anything they like with their arguments. Here are some of kind (* -> *): newtype Holds a = MkHolds a data Ignores a = MkIgnores newtype Wants a = MkWants (a -> Bool) data List a = MkPairList a (List a) | MkEmptyList data Maybe a = Just a | Nothing I don't think you can divide them up into two categories... -- Ashley Yakeley, Seattle WA

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by ��Ż�չЭ�� 06 Jun '02

06 Jun '02

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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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