- Haskell-Cafe - Haskell.org

RE: [Haskell-cafe] Ord (IORef a)?
by Simon Peyton-Jones 23 Jun '04

23 Jun '04

| From: John Meacham | would it be possible to provide an Ord instance for (IORef a)? For | things like loop detection, one may need to make many IORef comparasions | and being able to use an efficient set would be a really big win. | | Since IORefs are created in the IO monad, the actual order can be | arbitrary without breaking referential transparency (as long as it | doesn't change over the lifetime of the IORef.) I think that'd be entirely reasonable. Two runs of the same program might give observably different behaviour wrt Ord, but that's ok because the refs are allocates in the monad, as you say. There is an efficiency cost though. Each IORef would need to have an extra field, to record its allocation ID. (Address is not enough -- the garbage collector can mangle them.) My own view is that this is fine -- IORefs shouldn't be in your inner loop, so an extra word in each is no big deal. Unless there are views to the contrary, I'd be happy to see this in GHC. Simon

4 3

Re: [Haskell-cafe] Polymorphic algebraic type constructors
by MR K P SCHUPKE 22 Jun '04

22 Jun '04

>to be [] in each case. ahh but in this example: f :: [Int] -> [Bool] f (i:is) = even i : f is f e@[] = e e is an empty list of Ints not an empty list of Bools! you mean: f :: [Int] -> [Bool] f (i:is) = even i : f is f _ = [] Keean.

3 3

How to use QSem?
by S. Alexander Jacobson 22 Jun '04

22 Jun '04

The GHC documentation on QSem is very sparse. I would like to give a thread exclusive access to a resource. My *guess* based on the documentation is that I can create an exclusive lock using: logSem <- newQSem 1 And then any thread that wants to lock the resource uses: withLogSem x = do waitQSem logSem; y <- x; signalQSem logSem; return y as follows: withLogSem $ rotate curLogPos Am I misunderstanding QSem? -Alex- _________________________________________________________________ S. Alexander Jacobson mailto:me@alexjacobson.com tel:917-770-6565 http://alexjacobson.com

1 0

Re: [Haskell-cafe] Polymorphic algebraic type constructors
by MR K P SCHUPKE 22 Jun '04

22 Jun '04

>(I still think it's a bug though:-) It is definitely not a bug... you cannot assert that the types: Either String a Either String b are both equal ant not equal at the same time. You either mean: f :: (a->a) -> Either String a -> Either String a Or you mean f :: (a->b) -> Either String a -> Either String b actually the second form a and b can be the same but they cannot be unified inside the closure. One way of doing this is to group all the string types on one side: so instead of: data A a = A a | B String | C String | D String you do: data A a = A a | B Strings data Strings = S1 String | S2 String | S3 String you can then write: f :: (a->b) -> A a -> A b f g (A a) = (A $ g a) f g (B a) = (B a) and use (S1 . B), (S2 . B), (S3 . B) to access each string. Keean.

2 1

Re: [Haskell-cafe] Polymorphic algebraic type constructors
by MR K P SCHUPKE 22 Jun '04

22 Jun '04

>data E a b > = L1 { r :: a } > | L2 { r :: a } > | L3 { r :: a } > | L4 { r :: a } > | R b > deriving Show How is this different from: data E a b = L1 a | L2 a | R b f g (R a) = R (g a) f _ other = other Isn't this more or less what was in the original... The problem is the type: f :: (a->b) -> Either String a -> Either String b this line: f _ other = other says: Either String a == Either String b -- which it doesn't Keean.

2 2

Re: [Haskell-cafe] Polymorphic algebraic type constructors
by MR K P SCHUPKE 22 Jun '04

22 Jun '04

>If I have a polymorphic algebraic type (T a) with several type data MyType a = A a | B String | C Int So how do you expect to get the contents of B and C outm when they are differenf Types - surely you mean: data MyType a = A a | B MyStrings Keean.

1 0

Ord (IORef a)?
by John Meacham 22 Jun '04

22 Jun '04

would it be possible to provide an Ord instance for (IORef a)? For things like loop detection, one may need to make many IORef comparasions and being able to use an efficient set would be a really big win. Since IORefs are created in the IO monad, the actual order can be arbitrary without breaking referential transparency (as long as it doesn't change over the lifetime of the IORef.) John -- John Meacham - ⑆repetae.net⑆john⑈

3 3

Using HaXml filters for XML namespaces
by Graham Klyne 21 Jun '04

21 Jun '04

I am (at last!) looking to implement an XML namespace processing filter for HaXml, having completed the various structural changes needed to carry namespace information in the content model. The basic idea is to process a document such that: (a) all xmlns:xx='xxuri' attributes are removed (b) all attribute and element names of the form xx:foo are replaced by a URI xxuri and local name foo, corresponding to a currently in-scope namespace declaration. This raises some questions, which may simply be that I'm overlooking some obvious way of using the CFilter combinator library: (1) is there a combinator for updating attributes, in a fashion similar to the way 'chip' updates elements? (I wrote my own "atip" for general entity substitution in attributes, but it seems that there should be a way to do this in the library.) (2) it appears that I need the XML processing filter to carry a symbol table of in-scope namespaces, which will be updated through the traversal as new xmlnx:xx declarations come into scope. I cannot see any way to do this with the existing combinators; I think this would need a form of recursive combinator, along the lines of 'foldXml' that uses a newly computed CFilter function for each successive recursion. I imagine something like: mangleXml f mkf c = (f' `o` (chip (mangleXml f` mkf))) c where f' = mkf f c (where I would use mkf to scan the attributes and create an augmented version of f that takes into account new xmlns declarations, and f would be a function that replaces names with qnames on an element and its attributes.) Am I missing a trick with the existing combinators? ... (An alternative approach I considered was to use a monadic form of CFilter function which could carry and update the namespace table as the traversal progresses, but that appears to require that (a) many of the existing combinators would have be re-written, and (b) it doesn't cleanly deal with issues of scoping xmlns declarations.) ... #g ------------ Graham Klyne For email: http://www.ninebynine.org/#Contact

1 0

Re: (to haskell-cafe) comment on language shootout
by Iavor S. Diatchki 21 Jun '04

21 Jun '04

hello, Greg Morrisett wrote: > Iavor S. Diatchki wrote: > >> what would be cool is to have a place where >> one has many versions of the same program, but each written in the >> "most natural" >> way for the particular language --- a kind of "rosetta stone" for >> programming languages. > > > The ICFP programming contest (and other contests) serve this > purpose pretty well. I've found it very instructive to stare > at solutions in languages with which I'm not familiar. And the > tasks at hand tend to be more realistic (e.g., rendering) > than micro-benchmarks. that's a good point! one thing i don't like about programming contests, is that usually time is very limited, so often one ends up with terrible hacks that kind of work... one might argue though, that "real programmers" are often faced with such situations. -iavor

2 1

Laws in category theory..
by Ron de Bruijn 21 Jun '04

21 Jun '04

I have been thinking about the following laws of a monad. o is composition. # associative law of a monad mu o (mu o T) == mu o (T o mu) The notation: Tmu, that's the same as T o mu, right? How do I relate this to x*(y*z) == (x*y)*z ? # the identity law of a monad mu o (T o eta) =={id}_\mathcal{C}= mu o (eta o T) I also don't understand why I can't create the following law out of the assocaitive law: (mu o T) == (T o mu) , bacause the first part of the expression is the same. Please, enlighten me :) Regards, Ron __________________________________ Do you Yahoo!? New and Improved Yahoo! Mail - 100MB free storage! http://promotions.yahoo.com/new_mail

2 1