Re: How can I make a counter without Monad?
Thanks for your help. Are there other ways to implement a counter in Haskell? Using a State monad? If I use your example on : test = let Node x l = enumeratedTree ( Node 'a' [undefined, Node 'b' []]) in tail l GHCI answers [Node (*** Exception: Prelude.undefined A monadic counter imposes an order of evaluation. In my program, I don't care about the order of the numbers. I only want them to be all different. I think a monad is too restrictive for what I need. From some of my code: let enumeratedTree = (`evalState` (0::Int)) $ (`mapTreeM` t) $ \x - do n - next return (n, x) next = do a - get; put $! succ a; return a where mapTreeM :: Monad m = (a - m b) - Tree a - m (Tree b) mapTreeM f (Node a ts) = do b - f a ts' - mapM (mapTreeM f) ts return (Node b ts') (which could also be an instance of a popular non-standard FunctorM class) Best regards Tomasz ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
RE: alpha problems with ghc 6.4
On 16 March 2005 04:14, Ian Lynagh wrote:
An alpha build of ghc 6.4 quickly fails because of the
#if alpha_TARGET_ARCH
import PrimRep ( getPrimRepSize, isFloatingRep )
import Type ( typePrimRep )
#endif
in ghc/compiler/typecheck/TcForeign.lhs which no longer exist.
Fortunately, the imported functions aren't used either.
Unfortunately, the build then fails when it comes to try to compile
this
same file as the typeMachRepRep function, used in this piece of code:
\begin{code}
#include nativeGen/NCG.h
#if alpha_TARGET_ARCH
checkFEDArgs arg_tys
= check (integral_args = 32) err
where
integral_args = sum [ machRepByteWidth rep
| (rep,hint) - map typeMachRepRep arg_tys,
hint /= FloatHint ]
err = ptext SLIT(On Alpha, I can only handle 4
non-floating-point arguments to foreign export dynamic) #else
checkFEDArgs arg_tys = returnM ()
#endif
\end{code}
doesn't exist. Is this fixable?
I think you want to use something like typeMachRep in
deSugar/DsForeign.lhs.
Cheers,
Simon
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Re: How can I make a counter without Monad?
On Wed, Mar 16, 2005 at 10:51:08AM +0100, Nicolas Oury wrote: Thanks for your help. Are there other ways to implement a counter in Haskell? Using a State monad? If I use your example on : test = let Node x l = enumeratedTree ( Node 'a' [undefined, Node 'b' []]) in tail l GHCI answers [Node (*** Exception: Prelude.undefined A monadic counter imposes an order of evaluation. In my program, I don't care about the order of the numbers. I only want them to be all different. I think a monad is too restrictive for what I need. OK, I understand. In this situation you probably want either splittable name supply. Let me get back to your first post... Best regards Tomasz ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: How can I make a counter without Monad?
On Wed, Mar 16, 2005 at 01:17:51AM +0100, Nicolas Oury wrote: * linear implicit parameters instance Splittable Int where split n = (2*n,2*n+1) But I have a problem : the counter value increases exponentially. (I can only count up to 32 elements...) Is there another way to split Int? You could use unbounded Integers, or forget about numbers and use lists of bits. newtype BitString = BitString [Bool] instance Splittable BitString where split (BitString bs) = (BitString (False : bs), BitString (True : bs)) Best regards Tomasz ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
RE: ghc-pkg too happy to create ~/.ghc
Thanks, I've committed a version of your patch. Cheers, Simon On 16 March 2005 04:07, Ian Lynagh wrote: The Debian autobuilders don't let you write to ~ (which seems reasonable, as they are only compiling the software, not running it), so my builds are failing with -- ==fptools== /usr/bin/make boot -wr; in /build/buildd/ghc6-6.4/ghc/rts [...] ../utils/ghc-pkg/ghc-pkg-inplace --force --update-package package.conf.inplace Creating user package database in /org/buildd/.ghc/sparc-linux-6.4/package.conf Fail: createDirectory: permission denied (Permission denied) -- The patch below fixes it. I'm not sure I understand why the code is written as it is, though. It looks to me like if any config file given by a FlagConfig is missing then the readFile in readParseDatabase is going to fall over. I don't know what should happen when modifying if there are -f options, so can't suggest a complete replacement. Thanks Ian --- ghc6-6.4.orig/ghc/utils/ghc-pkg/Main.hs +++ ghc6-6.4/ghc/utils/ghc-pkg/Main.hs @@ -269,10 +269,6 @@ archdir = appdir `joinFileName` subdir user_conf = archdir `joinFileName` package.conf b - doesFileExist user_conf - when (not b) $ do - putStrLn (Creating user package database in ++ user_conf) - createDirectoryIfMissing True archdir - writeFile user_conf emptyPackageConfig let -- The semantics here are slightly strange. If we are @@ -281,20 +277,23 @@ -- If we are not modifying (eg. list, describe etc.) then -- the user database is included by default. databases - | modify = foldl addDB [global_conf] flags - | not modify = foldl addDB [user_conf,global_conf] flags + | modify || not b = foldl addDB [global_conf] flags + | not modify = foldl addDB [user_conf,global_conf] flags -- implement the following rules: -- --user means overlap with the user database -- --global means reset to just the global database -- -f file means overlap with file - addDB dbs FlagUser = if user_conf `elem` dbs - then dbs - else user_conf : dbs + addDB dbs FlagUser +| (modify || b) (user_conf `notElem` dbs) = user_conf : dbs addDB dbs FlagGlobal = [global_conf] addDB dbs (FlagConfig f) = f : dbs addDB dbs _ = dbs + when (not b user_conf `elem` databases) $ do + putStrLn (Creating user package database in ++ user_conf) + createDirectoryIfMissing True archdir + writeFile user_conf emptyPackageConfig db_stack - mapM readParseDatabase databases return db_stack ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
RE: [Haskell] RE: ANNOUNCE: GHC version 6.4
On 15 March 2005 18:26, Sebastian Sylvan wrote: Ah, that did it! However, some packages seem to be missing, like HGL for instance. rts-1.0, base-1.0, haskell98-1.0, template-haskell-1.0, unix-1.0, Cabal-1.0, parsec-1.0, haskell-src-1.0, network-1.0, QuickCheck-1.0, HUnit-1.1, mtl-1.0, fgl-5.2, OpenGL-2.0, stm-1.0, readline-1.0, (lang-1.0), (concurrent-1.0), (posix-1.0), (util-1.0), (data-1.0), (text-1.0), (net-1.0), (hssource-1.0) How would one go about just installing every damn package availble? It looks like the configure machinery didn't detect the X11 libraries, or for some other reason failed to build the X11 package and hence also omitted the HGL package. Could you send the full ./configure output? Cheers, Simon ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: How can I make a counter without Monad?
Le 16 mars 05, à 11:08, Tomasz Zielonka a écrit : On Wed, Mar 16, 2005 at 01:17:51AM +0100, Nicolas Oury wrote: * linear implicit parameters instance Splittable Int where split n = (2*n,2*n+1) But I have a problem : the counter value increases exponentially. (I can only count up to 32 elements...) Is there another way to split Int? You could use unbounded Integers, or forget about numbers and use lists of bits. newtype BitString = BitString [Bool] instance Splittable BitString where split (BitString bs) = (BitString (False : bs), BitString (True : bs)) Best regards Tomasz OK, I have written instance Splittable Integer where split n = (2*n,2*n+1) foo::(%x::Integer) = [a] - [(a,Integer)] foo [] = [] foo (a:l) = (a,%x):(foo l) test = let %x = 1 in foo [1..15000] But, in this example, the numbering is linear and so test becomes quadratic. The main complexity of the program come from the numbering... (When you test it with ghci, this example is really slow) The same thing hapens with a list of bools. Best regards, Nicolas Oury ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: How can I make a counter without Monad?
On 2005-03-16 02:52:39 -0800, Nicolas Oury [EMAIL PROTECTED] said: instance Splittable Integer where split n = (2*n,2*n+1) foo::(%x::Integer) = [a] - [(a,Integer)] foo [] = [] foo (a:l) = (a,%x):(foo l) test = let %x = 1 in foo [1..15000] But, in this example, the numbering is linear and so test becomes quadratic. The main complexity of the program come from the numbering... (When you test it with ghci, this example is really slow) I haven't played much with the Splittable class yet, but what would be wrong with instance Splittable Integer where split n = (n,n+1) ? -- Peter Davis [EMAIL PROTECTED] Furthermore, I believe bacon prevents hair loss! ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
Re: How can I make a counter without Monad?
On Wed, Mar 16, 2005 at 10:51:08AM +0100, Nicolas Oury wrote: A monadic counter imposes an order of evaluation. In my program, I don't care about the order of the numbers. I only want them to be all different. I think a monad is too restrictive for what I need. This is a common misconception, there is nothing about monads that requires an order of evaluation. for instance none of Control.Monad.Identity - isomorphic to not using monads at all Control.Monad.Reader - distributes a value to subcomputations Control.Monad.Writer - collects values from subcomputations imply any particular order of evaluation. This is one of the major powers of Monads, they can encapsulate all sorts of 'side effects', not just order of evaluation or linear state. for this app, I would use Control.Monad.State or if I needed the extra lazyness, Control.Monad.Reader with an explicitly splittable namesupply. John -- John Meacham - repetae.netjohn ___ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
