Re: [Haskell-cafe] ANN: new version of uu-parsinglib
On Mon, Jun 01, 2009 at 08:27:05PM +0200, S. Doaitse Swierstra wrote: And rename empty to fail? You managed to confuse me since I always use pSucceed to recognise the empty string. That would clash with the existing and widely used fail. One could view empty as the parser for the empty language. And some of us are also interested in Alternatives that aren't parsers. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Trouble with types
Vladimir Reshetnikov wrote: Hi Daniel, Could you please explain what does mean 'monomorphic' in this context? I thought that all type variables in Haskell are implicitly universally quantified, so (a - a) is the same type as (forall a. a - a) At the top level (i.e. definition level), yes. However, each use site this polymorphism may be restricted down (in particular, to the point of monomorphism). In a syntax more like Core, the definition of the identity function is, id :: forall a. a - a id @a (x :: a) = x Where the @ is syntax for reifying types or for capital-lambda application (whichever interpretation you prefer). In this syntax it's clear to see that |id| (of type forall a. a - a) is quite different than any particular |id @a| (of type a-a for the particular @a). So in your example, there's a big difference between these definitions, (f,g) = (id,id) (f', g') @a = (id @a, id @a) The latter one is polymorphic, but it has interesting type sharing going on which precludes giving universally quantified types to f' and g' (the universality is for the pair (f',g') and so the types of f' and g' must covary). Whereas the former has both fields of the tuple being polymorphic (independently). Both interpretations of the original code are legitimate in theory, but the former is much easier to work with and reason about. It also allows for tupling definitions as a way of defining local name scope blocks, without side effects on types. -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Possible Haskell Project
Tom Hawkins wrote: At the core, the fundamental problem is not that complicated. It's just storing and retrieving a person's various health events: checkups, prescriptions, procedures, test results, etc. The main technical challenges are database distribution and patient security. Both are fun problems, and our friends at Galios continue to show how effective Haskell is at building secure systems. Any thoughts? Ideas? Actually, it's a lot more complicated than that, albeit not for technical reasons. There's a great deal of legislation about what can and cannot be done with medical records: who can have access to them, under what circumstances, how they can be transmitted, stored, etc. This is more than just boilerplate code--- clinics can be audited to prove proper handling and can loose their licenses or worse for improper handling of records. Additionally, the requisite formats do require a lot of boilerplate code since the protocols were defined back in the paper age and medical legislation moves at the speed of mountains. I worked briefly on an open-source database project for managing a medical clinic's records (so, not even for dealing with the public in any way). The technical feat isn't that difficult, as you say, but the human engineering involved can be quite complex--- and the human programming will have major effects on the design, in order to forbid invalid or unacceptable behavior. It's not a project to undertake lightly or without corporate funding. Medical record management is a market that has very low penetration from the F/OSS movement, which in turn places a burden on smaller clinics, so I'm all for anyone who's willing to invest in an open solution :) -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Checking a value against a passed-in constructor?
Hi Richard, Yeek. Why do you want to do _that_? Heh. I've got a parser and I want to check what I've parsed (it's an exercise in Write Yourself a Scheme in 48 Hours). check (Atom _) (Atom _) = True check (Bool _) (Bool _) = True check __= False Yes I came up with this too, but it seemed ugly to create unnecessary new values just to take them apart again. is_atom (Atom _) = True is_atom _= False This is nicer. It still requires listing out the possible constructors (Bool, Atom ... the real code obviously has more). I don't like that, because I've already listed them out once, in the type declaration itself. Surely, I shouldn't have to list them out again? There are various meta-programming (Scrap Your Boilerplate, Template Haskell) approaches you can use to automate some of these. You hit the nail on the head. Why I am doing this is because of boilerplate. Boilerplate gives me rashes and bulbous spots on the nose. Consider the following Ruby code: def check(zeClass, zeValue) zeValue.is_a? zeClass end This does not require a new function for every class defined in Ruby. (To be fair, though, the class of a Ruby object tells you precious little, compared to a Haskell type constructor). I figured there would be a clever Haskell idiom that would give me a similarly concise route. Does it really require Template Haskell? I can barely parse regular Haskell as it is.. Cheers, - Dan ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] HaL4: Haskell-Meeting in Germany, 12th June 2009
Hi all, If you are anyway near Halle/Saale in June, be sure not to miss out on: http://iba-cg.de/hal4.html We have already close to 50 registered participants, so expect a very lively meeting. See you there? (Late registration still possible.) Ciao, Janis. -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:vo...@tcs.inf.tu-dresden.de ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] HaL4: Haskell-Meeting in Germany, 12th June 2009
Janis Voigtlaender wrote: Hi all, If you are anyway near Halle/Saale in June, be sure not to miss out on: I meant anywhere near, of course :-) And even if you are not anyway or anywhere near, you might still want to come just for the occasion :-) -- Dr. Janis Voigtlaender http://wwwtcs.inf.tu-dresden.de/~voigt/ mailto:vo...@tcs.inf.tu-dresden.de ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Compiling a windows app - embedding application icon
Hi all, is it possible to make ghc embedd an application icon in the .exe during the compilation process? Günther ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] HXT XmlPicklers - TH Derivation
Hi, I have developed some simple TH code to automatically derive XmlPickler instances for my types and if there is interest, I will clean it up and submit a patch. Its not complete, but is a start. Any interest? Max ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Spam apology
Sorry for all the repeated messages, my e-mail client exploded. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Compiling a windows app - embedding application icon
Hello Gu?nther, Tuesday, June 2, 2009, 4:47:55 PM, you wrote: is it possible to make ghc embedd an application icon in the .exe during the compilation process? i've found that answer may be googled as gcc icon: 1) create icon.rc containing one line: 100 ICON freearc.ico 2) compile it using windres: windres.exe icon.rc icon.o 3) link in icon.o when making executable: ghc --make main.hs icon.o -- Best regards, Bulatmailto:bulat.zigans...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Possible Haskell Project
On Tue, Jun 2, 2009 at 12:12 AM, Antoine Latteraslat...@gmail.com wrote: A good place to start is http://en.wikipedia.org/wiki/HL7 , which is a not-for-profit organization which tries to define interfacing standards between medical devices and medical records providers. I haven't worked much with their standards so I don't know how useful they'd be. I think they might be geared towards vendor-to-vendor interop. As for the legacy of people who thought it wasn't complex: http://histalk.blog-city.com/guest_article__repeat_after_me_healthcare_data_models_matter.htm I don't agree with everything the guy wrote, but it's an interesting article. In an industry like this that generates so much data, I think all parties are tempted to record and track as must as possible. But after all the lab results, x-rays, and MRIs, it's the two or three paragraphs of a doctor's dictation that matter. Maybe patients and doctors would be best served if they had an easy way to store, retrieve, and query these dictations. I see this as an abstracting database problem: - records (dictations) are write-once-read-only data pertaining to a subject (patient) - some users (doctors, patients) are allowed to view a subset of records on a subject - some users are allowed to create new records on a subject - some users are allowed to change capabilities of other users Then, built on top of an abstract distributed data storage problem: - a network of computers store a collection of write-once-read-only data chunks (encoded, fragmented records) - chunks are distributed to minimize access time - chunks are distributed to maintain data integrity through system failures Both of these abstract problems can be used for many things outside the medical field. So even if an electronic health record project does not pan out, the code could find its way into other applications. -Tom ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Possible Haskell Project
The Dutch government has been trying to get something like this for years; parliament is asking every new minister why the promised heaven has not yet arrived, only to hear that more consultants are needed. I have been to hearings of our parliament and I can tell you such events are extremely informative and make you loose any hope that something good will come out of this soon; there are just too many stakeholders, and no so-called problem-owners except you. Simple questions asked, for which there often is no answer is: - who owns the information? - are you allowed to change information which you own? - should docters pay for the right to enter information in this system, or be paid for the service they provide if they enter information? Instead of trying to change the world you may run a small wiki (I run Twiki) server on your home machine where you just store the information you collect, and enter your information while you are having a consult through your iPhone! When you leave the room you ask your docter whether what you have entered is a correct view of this situation ;-}I think this will solve the major part of your problem, and maybe it opens the eyes of the medical establishment. Doaitse Swierstra On 2 jun 2009, at 11:18, wren ng thornton wrote: Tom Hawkins wrote: At the core, the fundamental problem is not that complicated. It's just storing and retrieving a person's various health events: checkups, prescriptions, procedures, test results, etc. The main technical challenges are database distribution and patient security. Both are fun problems, and our friends at Galios continue to show how effective Haskell is at building secure systems. Any thoughts? Ideas? Actually, it's a lot more complicated than that, albeit not for technical reasons. There's a great deal of legislation about what can and cannot be done with medical records: who can have access to them, under what circumstances, how they can be transmitted, stored, etc. This is more than just boilerplate code--- clinics can be audited to prove proper handling and can loose their licenses or worse for improper handling of records. Additionally, the requisite formats do require a lot of boilerplate code since the protocols were defined back in the paper age and medical legislation moves at the speed of mountains. I worked briefly on an open-source database project for managing a medical clinic's records (so, not even for dealing with the public in any way). The technical feat isn't that difficult, as you say, but the human engineering involved can be quite complex--- and the human programming will have major effects on the design, in order to forbid invalid or unacceptable behavior. It's not a project to undertake lightly or without corporate funding. Medical record management is a market that has very low penetration from the F/OSS movement, which in turn places a burden on smaller clinics, so I'm all for anyone who's willing to invest in an open solution :) -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Success and one last issue with Data.Binary
I've got the following printHex string as a response from a 9P server
running on the Inferno Operating System. (thanks to a friendly mailing list
contributor who sent a nice example of using Data.Binary)
13006500040600395032303030
This is a little endian encoded ByteString with the following fields in it:
Rversion {size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
But when I try to use the following implementation of get to decode this
stream, I'm getting the following error:
too few bytes. Failed reading at byte position 20
Unfortunately, I'm only expecting 19 bytes, and in fact never asked for byte
20. (I am just asking for everything up to ssize, and then
getRemainingLazyByteString).
Is this a bug?Is it mine or in Data.Binary? :-)
Here's my get function:
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
The good news is I'm talking 9P otherwise, correctly, just having some
decoding issues. I hope to have a hackage package eventually for this.
Dave
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
I think getRemainingLazyByteString expects at least one byte (this,
perhaps, is not the appropriate behavior). You'll want to wrap your
call to getRemainingLazyByteString with a call to
Data.Binary.Get.remaining[1] like this:
foo = do
r - remaining
rbs - case r of
0 - return empty -- Data.ByteString.Lazy.empty
_ - getRemainingLazyByteString
Hope this helps. :)
/jve
1:
http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3Aremaining
On Tue, Jun 2, 2009 at 12:20 PM, David Leimbach leim...@gmail.com wrote:
I've got the following printHex string as a response from a 9P server
running on the Inferno Operating System. (thanks to a friendly mailing list
contributor who sent a nice example of using Data.Binary)
13006500040600395032303030
This is a little endian encoded ByteString with the following fields in it:
Rversion {size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
But when I try to use the following implementation of get to decode this
stream, I'm getting the following error:
too few bytes. Failed reading at byte position 20
Unfortunately, I'm only expecting 19 bytes, and in fact never asked for byte
20. (I am just asking for everything up to ssize, and then
getRemainingLazyByteString).
Is this a bug? Is it mine or in Data.Binary? :-)
Here's my get function:
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
The good news is I'm talking 9P otherwise, correctly, just having some
decoding issues. I hope to have a hackage package eventually for this.
Dave
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--
/jve
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary 0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag:: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
Thomas,
You're correct. For some reason, I based my advice on the thought that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary 0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
--
/jve
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Re: [Haskell-cafe] Checking a value against a passed-in constructor?
On Tue, Jun 2, 2009 at 3:50 AM, Dan danielkc...@gmail.com wrote: You hit the nail on the head. Why I am doing this is because of boilerplate. Boilerplate gives me rashes and bulbous spots on the nose. Consider the following Ruby code: def check(zeClass, zeValue) zeValue.is_a? zeClass end This does not require a new function for every class defined in Ruby. (To be fair, though, the class of a Ruby object tells you precious little, compared to a Haskell type constructor). I figured there would be a clever Haskell idiom that would give me a similarly concise route. Does it really require Template Haskell? I can barely parse regular Haskell as it is.. So the question is, why do you need to know if x is an Atom or a Bool? The Haskell idiom is to pattern match and just do what you want with the data: f (Atom s) = ... f (Bool b) = ... instead of f x = if isAtom x then ... atomData x ... else ... boolData x ... Alternatively, you can define a fold[1] once: myval :: MyVal - (Bool - a) - (String - a) - a myval (Bool b) bool atom = bool b myval (Atom s) bool atom = atom s f x = myval bool atom where bool b = ... atom s = ... This is a small amount of boilerplate that you write once for each type; it's possible to automate it with TH, but usually it's not worth it, in my opinion. Coming from Ruby (the same route I took to get to Haskell!), you should be aware that Haskell does have somewhat more boilerplate than Ruby, but it has its benefits as well. I am a convert to the Church of Purity and Type-Safety :) And you can use type classes for many metaprogramming tasks. -- ryan [1] fold here is the general term for this type of function. Examples are foldr: http://haskell.org/ghc/docs/latest/html/libraries/base/src/GHC-Base.html#foldr maybe: http://haskell.org/ghc/docs/latest/html/libraries/base/src/Data-Maybe.html#maybe either: http://haskell.org/ghc/docs/latest/html/libraries/base/src/Data-Either.html#either ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Compiling a windows app - embedding a manifest
Hi all, is it possible to make ghc embedd a particular manifest in the .exe during the compilation process? Günther ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Success and one last issue with Data.Binary
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type for a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm sending.
The strings are *NOT* null terminated in 9p, and this is little endian for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6 bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so why am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag:: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
--
/jve
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type for a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm sending.
The strings are *NOT* null terminated in 9p, and this is little endian for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6 bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so why am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
--
/jve
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 10:24 AM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
I've shown that I am not trying to decode more than I'm providing. I've
asked, expliciitly, for 13 bytes, and then remaining, and the library is
complaining about the 20th byte.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
The other branch is Rerror, which is a shorter message decode stream.
Unfortunately, I can't get Debug.Trace to show anything to prove it's
taking this fork of the code. I suppose I could unsafePerformIO :-)
Perhaps I just need a new version of binary?? I'll give it a go and try
your version. But I need to decode over a dozen message types, so I will
need a case or guard or something.
Dave
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag:: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $ Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian for
19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6 bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag:: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
Re: [Haskell-cafe] Success and one last issue with Data.Binary
Perhaps there's some place in your code that's forcing the lazy read
to consume more. Perhaps you could replace it with an explict (and
strict) getBytes[1] in combination with remaining[2]?
Is there a reason you want to use lazy byte strings rather than
forcing full consumption? Do the 9P packets generally have a lot of
trailing useless data?
1.
http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3AgetBytes
2.
http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3Aremaining
On Tue, Jun 2, 2009 at 4:28 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 10:24 AM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
I've shown that I am not trying to decode more than I'm providing. I've
asked, expliciitly, for 13 bytes, and then remaining, and the library is
complaining about the 20th byte.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
The other branch is Rerror, which is a shorter message decode stream.
Unfortunately, I can't get Debug.Trace to show anything to prove it's
taking this fork of the code. I suppose I could unsafePerformIO :-)
Perhaps I just need a new version of binary?? I'll give it a go and try
your version. But I need to decode over a dozen message types, so I will
need a case or guard or something.
Dave
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
--
/jve
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
What happens if you use `getRemainingLazyByteString' in your error
branch instead of `getLazyByteString'?
On Tue, Jun 2, 2009 at 4:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $ Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6 bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error
message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the
code
(which did change behavior due to the guard and data declarations
that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
[Haskell-cafe] code reviewers wanted for hashed-storage (darcs)
Dear Haskellers, Will you have a few spare hours this summer? The Darcs team needs your help! Summary --- We need two volunteers to help us review the standalone hashed-storage module, which will be used by Darcs in the future. Background -- Darcs supports 'hashed' repositories in which each file in the pristine cache is associated with a cryptographic hash. Hashed repositories help Darcs to resist some forms of corruption and also allow for nice features such as a global patch cache and lazy patch fetching. Unfortunately, our implementation can be rather slow. Petr has some nice ideas for making these repositories work a lot faster. He will be producing a library he calls 'hashed-storage' which generalises the idea of storing files associating them with a cryptographic hash and furthermore improves on the current implementation used by Darcs. The hashed-storage library is general purpose and may find a use in other applications that need to manage a large number of files. I'm excited about this project because if it succeeds, we can finally start hammering home the point that yes, Darcs *can* be a fast revision control system. It may not fix all our problems -- we'll still need a Darcs 3 -- but it should alleviate at least some of the practical day to day issues. I want to ensure that Petr's project is a success. One thing in particular is that I would like him to get regular code review from the Haskell community, perhaps from folks who aren't already on the Darcs team. Do you think you can help? Two volunteers needed - I'm looking for just two volunteers this summer: 1. Do you have one hour per week this summer? We need somebody to track and make comments on changes to hashed-storage over this summer. The current library has only 8 modules, with less than 1400 lines of code, so it should be relatively easy to get started :-) 2. Do you have a spare weekend in August? Towards the end of the summer, it would be nice to have somebody examine the final version of hashed-storage and give us their thoughts. Again, no Darcs experience is needed, as this is a standalone module. In fact, a fresh perspective would be very helpful. More information You can download hashed-storage from hackage from his darcs repository: darcs get http://repos.mornfall.net/hashed-storage For more information about his project, see http://web.mornfall.net/blog/summer_of_code.html Please get in touch with me if you can help. Thanks! Eric -- Eric Kow http://www.nltg.brighton.ac.uk/home/Eric.Kow PGP Key ID: 08AC04F9 pgphUQDmGStl4.pgp Description: PGP signature ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Compiling a windows app - embedding a manifest
Hello Gu?nther, Wednesday, June 3, 2009, 12:11:15 AM, you wrote: Hi all, is it possible to make ghc embedd a particular manifest in the .exe during the compilation process? add to .rc file: 1 24 app.manifest and put manifect into app.manifest -- Best regards, Bulatmailto:bulat.zigans...@gmail.com ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 1:32 PM, John Van Enk vane...@gmail.com wrote:
Perhaps there's some place in your code that's forcing the lazy read
to consume more. Perhaps you could replace it with an explict (and
strict) getBytes[1] in combination with remaining[2]?
Unfortunately, I'm using a Lazy ByteString network IO lib. So I don't think
going to a strict ByteString is going to be possible.
Is there a reason you want to use lazy byte strings rather than
forcing full consumption? Do the 9P packets generally have a lot of
trailing useless data?
Nope. Just I noticed that there was a Network ByteString package that
utilized lazy bytestrings :-).
Even if that's why it's going for a 20th byte, shouldn't that be a bug? :-)
1.
http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3AgetBytes
2.
http://hackage.haskell.org/packages/archive/binary/0.5.0.1/doc/html/Data-Binary-Get.html#v%3Aremaining
On Tue, Jun 2, 2009 at 4:28 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 10:24 AM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than providing.
I've shown that I am not trying to decode more than I'm providing. I've
asked, expliciitly, for 13 bytes, and then remaining, and the library
is
complaining about the 20th byte.
First check the length of the bytestring you pass in to the to level
decode (or 'get') routine and walk though that to figure out how much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us is
the branch being taken.
The other branch is Rerror, which is a shorter message decode stream.
Unfortunately, I can't get Debug.Trace to show anything to prove it's
taking this fork of the code. I suppose I could unsafePerformIO :-)
Perhaps I just need a new version of binary?? I'll give it a go and
try
your version. But I need to decode over a dozen message types, so I will
need a case or guard or something.
Dave
I think the issue isn't with the code provided. I cleaned up the code
(which did change behavior due to the guard and data declarations that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag:: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v }
put _ = undefined
--
/jve
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Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 1:36 PM, John Van Enk vane...@gmail.com wrote:
What happens if you use `getRemainingLazyByteString' in your error
branch instead of `getLazyByteString'?
I actually am using getRemainingLazyByteString right now, and it still
thinks I'm asking for a 20th byte.
if I delete the other guarded branch of that function, it still fails saying
I'm asking for the 20th byte.
Dave
On Tue, Jun 2, 2009 at 4:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type
for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little
endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6
bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so
why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error
message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com
wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought
that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with binary
0.5.0.1.
The specific error means you are requiring more data than
providing.
First check the length of the bytestring you pass in to the to
level
decode (or 'get') routine and walk though that to figure out how
much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us
is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the
code
(which did change behavior due to the guard and data declarations
that
weren't in the mailling) and it works fine all the way down to the
expected minimum of 13 bytes.
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag
Re: [Haskell-cafe] Success and one last issue with Data.Binary
Again, I can't reproduce your problem. Are you getting data through
some previous Binary instance before calling the routines you show us
here? The code I tested with is below - I've tried it with both
'getSpecific' paths by commenting out one path at a time. Both
methods work, shown below.
Thomas
*Main decode test :: RV
Rversion {size = 19, mtype = 101, tag = 65535, msize = 1024, ssize =
6, version = Chunk 9P2000 Empty}
*Main :q
Leaving GHCi.
[... edit ...]
[1 of 1] Compiling Main ( p.hs, interpreted )
Ok, modules loaded: Main.
*Main decode test :: RV
Rerror {size = 19, mtype = 101, tag = 65535, ssize = 1024, ename =
Chunk \NUL\NUL\ACK\NUL9P2000 Empty}
*Main
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size:: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
| Rerror { size:: Word32,
mtype :: Word8,
tag :: Word16,
ssize :: Word16,
ename :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
put = undefined
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
{- = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
-}
= do t - getWord16le
ss - getWord16le
e - getLazyByteString $ fromIntegral ss
return $ Rerror {size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
test = pack
[ 0x13
, 0x00
, 0x00
, 0x00
, 0x65
, 0xff
, 0xff
, 0x00
, 0x04
, 0x00
, 0x00
, 0x06
, 0x00
, 0x39
, 0x50
, 0x32
, 0x30
, 0x30
, 0x30 ]
On Tue, Jun 2, 2009 at 1:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $ Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6 bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
Re: [Haskell-cafe] Success and one last issue with Data.Binary
Just so we know that it's not the issue, what version of binary are
you using? The most current one is 0.5.0.1.
On Tue, Jun 2, 2009 at 4:46 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:36 PM, John Van Enk vane...@gmail.com wrote:
What happens if you use `getRemainingLazyByteString' in your error
branch instead of `getLazyByteString'?
I actually am using getRemainingLazyByteString right now, and it still
thinks I'm asking for a 20th byte.
if I delete the other guarded branch of that function, it still fails saying
I'm asking for the 20th byte.
Dave
On Tue, Jun 2, 2009 at 4:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type
for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little
endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6
bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so
why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le -- 4
ss - getWord16le -- 2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $ MessageClient
$
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error
message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com
wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought
that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with
binary
0.5.0.1.
The specific error means you are requiring more data than
providing.
First check the length of the bytestring you pass in to the to
level
decode (or 'get') routine and walk though that to figure out how
much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave us
is
the branch being taken.
I think the issue isn't with the code provided. I cleaned up the
code
(which did change behavior due to the guard and data declarations
that
weren't in the mailling) and it works fine all the way down to the
expected
Re: [Haskell-cafe] Success and one last issue with Data.Binary
0.5.0.1
On Tue, Jun 2, 2009 at 1:56 PM, John Van Enk vane...@gmail.com wrote:
Just so we know that it's not the issue, what version of binary are
you using? The most current one is 0.5.0.1.
On Tue, Jun 2, 2009 at 4:46 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:36 PM, John Van Enk vane...@gmail.com wrote:
What happens if you use `getRemainingLazyByteString' in your error
branch instead of `getLazyByteString'?
I actually am using getRemainingLazyByteString right now, and it still
thinks I'm asking for a 20th byte.
if I delete the other guarded branch of that function, it still fails
saying
I'm asking for the 20th byte.
Dave
On Tue, Jun 2, 2009 at 4:31 PM, David Leimbach leim...@gmail.com
wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com
wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little
endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response
type
for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit value for the length of the string I'm
sending.
The strings are *NOT* null terminated in 9p, and this is little
endian
for
6 bytes remaining.
5032303030 = 6 bytes the ASCII or UTF-8 string 9P2000 which is 6
bytes
4 + 1 + 2 + 4 + 2 + 6 = 19 bytes.
As far as I can see, my get code does NOT ask for a 20th byte, so
why
am I
getting that error?
get = do s - getWord32le -- 4
mtype - getWord8 -- 1
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le -- 2
ms - getWord32le --
4
ss - getWord16le --
2
v -
getRemainingLazyByteString -- remaining should be 6 bytes.
return $
MessageClient
$
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
Should I file a bug? I don't believe I should be seeing an error
message
claiming a failure at the 20th byte when I've never asked for one.
Dave
On Tue, Jun 2, 2009 at 10:51 AM, John Van Enk vane...@gmail.com
wrote:
Thomas,
You're correct. For some reason, I based my advice on the thought
that
19 was the minimum size instead of 13.
On Tue, Jun 2, 2009 at 1:24 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
I think getRemainingLazyByteString expects at least one byte
No, it works with an empty bytestring. Or, my tests do with
binary
0.5.0.1.
The specific error means you are requiring more data than
providing.
First check the length of the bytestring you pass in to the to
level
decode (or 'get') routine and walk though that to figure out how
much
it should be consuming. I notice you have a guard on the
'getSpecific' function, hopefully you're sure the case you gave
us
is
the branch being taken.
I
Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 1:56 PM, Thomas DuBuisson thomas.dubuis...@gmail.com
wrote:
Again, I can't reproduce your problem. Are you getting data through
some previous Binary instance before calling the routines you show us
here?
Ah good question... I'm calling decode, but it's not clear that it's even
running my instance of Get
If I have a lazy bytestring, and call decode, which instance of Get
runs? Probably not my 9P message version I'll bet...
geeze... :-(
The code I tested with is below - I've tried it with both
'getSpecific' paths by commenting out one path at a time. Both
methods work, shown below.
Thomas
*Main decode test :: RV
Rversion {size = 19, mtype = 101, tag = 65535, msize = 1024, ssize =
6, version = Chunk 9P2000 Empty}
*Main :q
Leaving GHCi.
[... edit ...]
[1 of 1] Compiling Main ( p.hs, interpreted )
Ok, modules loaded: Main.
*Main decode test :: RV
Rerror {size = 19, mtype = 101, tag = 65535, ssize = 1024, ename =
Chunk \NUL\NUL\ACK\NUL9P2000 Empty}
*Main
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size:: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
| Rerror { size:: Word32,
mtype :: Word8,
tag :: Word16,
ssize :: Word16,
ename :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
put = undefined
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
{- = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
-}
= do t - getWord16le
ss - getWord16le
e - getLazyByteString $ fromIntegral ss
return $ Rerror {size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
test = pack
[ 0x13
, 0x00
, 0x00
, 0x00
, 0x65
, 0xff
, 0xff
, 0x00
, 0x04
, 0x00
, 0x00
, 0x06
, 0x00
, 0x39
, 0x50
, 0x32
, 0x30
, 0x30
, 0x30 ]
On Tue, Jun 2, 2009 at 1:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
65 = 1 byte for message type. 65 is Rversion or the response type
for
a
Tversion request
= 2 bytes for 16bit message tag.
0004 = 4 bytes for the 32 bit maximum message payload size I'm
negotiating with the 9P server. This is little endian for 1024
0600 = 2 bytes for 16 bit
Re: [Haskell-cafe] Success and one last issue with Data.Binary
It will run the instance of the inferred type (or you can provide a
type signature to force it). I've done this often before with lists -
trying to read in some arbitrary, typically high, number of elements
causes issues :-)
Thomas
On Tue, Jun 2, 2009 at 2:07 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:56 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
Again, I can't reproduce your problem. Are you getting data through
some previous Binary instance before calling the routines you show us
here?
Ah good question... I'm calling decode, but it's not clear that it's even
running my instance of Get
If I have a lazy bytestring, and call decode, which instance of Get
runs? Probably not my 9P message version I'll bet...
geeze... :-(
The code I tested with is below - I've tried it with both
'getSpecific' paths by commenting out one path at a time. Both
methods work, shown below.
Thomas
*Main decode test :: RV
Rversion {size = 19, mtype = 101, tag = 65535, msize = 1024, ssize =
6, version = Chunk 9P2000 Empty}
*Main :q
Leaving GHCi.
[... edit ...]
[1 of 1] Compiling Main ( p.hs, interpreted )
Ok, modules loaded: Main.
*Main decode test :: RV
Rerror {size = 19, mtype = 101, tag = 65535, ssize = 1024, ename =
Chunk \NUL\NUL\ACK\NUL9P2000 Empty}
*Main
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size :: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
| Rerror { size :: Word32,
mtype :: Word8,
tag :: Word16,
ssize :: Word16,
ename :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
put = undefined
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
{- = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
-}
= do t - getWord16le
ss - getWord16le
e - getLazyByteString $ fromIntegral ss
return $ Rerror {size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
test = pack
[ 0x13
, 0x00
, 0x00
, 0x00
, 0x65
, 0xff
, 0xff
, 0x00
, 0x04
, 0x00
, 0x00
, 0x06
, 0x00
, 0x39
, 0x50
, 0x32
, 0x30
, 0x30
, 0x30 ]
On Tue, Jun 2, 2009 at 1:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for 32bit message payload, This is little endian
for 19
bytes total.
Re: [Haskell-cafe] Success and one last issue with Data.Binary
On Tue, Jun 2, 2009 at 2:07 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:56 PM, Thomas DuBuisson
thomas.dubuis...@gmail.com wrote:
Again, I can't reproduce your problem. Are you getting data through
some previous Binary instance before calling the routines you show us
here?
Ah good question... I'm calling decode, but it's not clear that it's even
running my instance of Get
If I have a lazy bytestring, and call decode, which instance of Get
runs? Probably not my 9P message version I'll bet...
geeze... :-(
ANd... that was it. I totally didn't decode with the right decoder. By
the expression I had, it appears it was trying to decode a ByteString as a
String, and that was causing a big darned mess.
Thanks for all the help guys. I'm glad it's not a bug in the library, just
my dumb code
Dave
The code I tested with is below - I've tried it with both
'getSpecific' paths by commenting out one path at a time. Both
methods work, shown below.
Thomas
*Main decode test :: RV
Rversion {size = 19, mtype = 101, tag = 65535, msize = 1024, ssize =
6, version = Chunk 9P2000 Empty}
*Main :q
Leaving GHCi.
[... edit ...]
[1 of 1] Compiling Main ( p.hs, interpreted )
Ok, modules loaded: Main.
*Main decode test :: RV
Rerror {size = 19, mtype = 101, tag = 65535, ssize = 1024, ename =
Chunk \NUL\NUL\ACK\NUL9P2000 Empty}
*Main
import Data.ByteString.Lazy
import Data.Binary
import Data.Binary.Get
data RV =
Rversion { size:: Word32,
mtype :: Word8,
tag :: Word16,
msize :: Word32,
ssize :: Word16,
version :: ByteString}
| Rerror { size:: Word32,
mtype :: Word8,
tag :: Word16,
ssize :: Word16,
ename :: ByteString}
deriving (Eq, Ord, Show)
instance Binary RV where
put = undefined
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
{- = do t - getWord16le
ms - getWord32le
ss - getWord16le
v - getRemainingLazyByteString
return $ Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
-}
= do t - getWord16le
ss - getWord16le
e - getLazyByteString $ fromIntegral ss
return $ Rerror {size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
test = pack
[ 0x13
, 0x00
, 0x00
, 0x00
, 0x65
, 0xff
, 0xff
, 0x00
, 0x04
, 0x00
, 0x00
, 0x06
, 0x00
, 0x39
, 0x50
, 0x32
, 0x30
, 0x30
, 0x30 ]
On Tue, Jun 2, 2009 at 1:31 PM, David Leimbach leim...@gmail.com wrote:
On Tue, Jun 2, 2009 at 1:28 PM, John Van Enk vane...@gmail.com wrote:
I think Thomas' point was that some other branch in `getSpecific' is
running. Is there a chance we can see the rest of `getSpecific'?
Sure: (In the meantime, I'll try the suggested code from before)
get = do s - getWord32le
mtype - getWord8
getSpecific s mtype
where
getSpecific s mt
| mt == mtRversion = do t - getWord16le
ms - getWord32le
ss - getWord16le
v -
getRemainingLazyByteString
return $ MessageClient $
Rversion {size=s,
mtype=mt,
tag=t,
msize=ms,
ssize=ss,
version=v}
| mt == mtRerror = do t - getWord16le
ss - getWord16le
e - getLazyByteString $
fromIntegral ss
return $ MessageClient $
Rerror
{size=s,
mtype=mt,
tag=t,
ssize=ss,
ename=e}
On Tue, Jun 2, 2009 at 4:20 PM, David Leimbach leim...@gmail.com
wrote:
The thing is I have 19 bytes in the hex string I provided:
13006500040600395032303030
That's 38 characters or 19 bytes.
The last 4 are 9P2000
1300 = 4 bytes for
[Haskell-cafe] ANN: Anglohaskell 2009
Anglohaskell 2009 is go! I'm taking on the mantle of organiser, and Microsoft Research have offered us space for talks in Cambridge again. The event will be held on the 7th and 8th of August. More info at http://www.haskell.org/haskellwiki/AngloHaskell/2009 , planning and discussion in #anglohaskell on freenode. For those not familiar with Anglohaskell, it's a somewhat-informal get-together featuring a mixture of talks, discussion and socialising with topics from the hobbyist to the pragmatic to the theoretical. All are welcome, regardless of experience, and best of all - it's free! If anyone wants to offer a talk, help with running the event, accomodation for haskellers from out of town or some ideas, please feel free to edit the wiki page appropriately and/or give us a yell in #anglohaskell. -- Philippa Cowderoy fli...@flippac.org ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] ANNOUNCE: The Haskell Platform 2009.2.0.1
We're pleased to announce the second release of the Haskell Platform: a single, standard Haskell distribution for everyone. The specification, along with installers (including Windows and Unix installers for a full Haskell environment) are available. Download the Haskell Platform 2009.2.0.1: http://hackage.haskell.org/platform/ The Haskell Platform is a blessed library and tool suite for Haskell distilled from Hackage, along with installers for a wide variety of systems. It saves developers work picking and choosing the best Haskell libraries and tools to use for a task. Distro maintainers that support the Haskell Platform can be confident they're fully supporting Haskell as the developers intend it. Developers targetting the platform can be confident they have a trusted base of code to work with. What you get: http://hackage.haskell.org/platform/contents.html With regular time-based releases, we expect the platform will grow into a rich, indispensable development environment for all Haskell projects. Please note that this is a beta release. We do not expect all the installers to work perfectly, nor every developer need met, and we would appreciate feedback. You can help out by packaging the platform for your distro, or reporting bugs and feature requests, or installing Haskell onto your friends' machines. The process for adding new tools and libraries will be outlined in coming weeks. The Haskell Platform would not have been possible without the hard work of the Cabal development team, the Hackage developers and maintainers, the individual compiler, tool and library authors who contributed to the suite, and the distro maintainers who build and distribute the Haskell Platform. Thanks! -- The Platform Infrastructure Team ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ANN: Anglohaskell 2009
On Tue, 02 Jun 2009 23:45:18 +0200, Philippa Cowderoy fli...@flippac.org wrote: Anglohaskell 2009 is go! F.A.B. :) -- Regards, Henk-Jan van Tuyl -- http://functor.bamikanarie.com http://Van.Tuyl.eu/ -- ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] beginners question about fromMaybe
hi there heres a code snipped, don't care about the parameters. the thing is i make a lookup on my map m and then branch on that return value probePhase is sc [] m = [] probePhase is sc (x:xs) m | val == Nothing = probePhase is sc xs m | otherwise = jr ++ probePhase is sc xs m where jr = joinTuples sc x (fromMaybe [] val) key = getPartialTuple is x val = Map.lookup key m the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly because i know that it is not Nothing. is there a better way to solve this? regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
If you're absolutely certain that the lookup always succeeds, then you can use pattern matching as follows: where jr = joinTuples sc x val key = getPartialTuple is x Just val = Map.lookup key m On 6/3/09, Nico Rolle nro...@web.de wrote: hi there heres a code snipped, don't care about the parameters. the thing is i make a lookup on my map m and then branch on that return value probePhase is sc [] m = [] probePhase is sc (x:xs) m | val == Nothing = probePhase is sc xs m | otherwise = jr ++ probePhase is sc xs m where jr = joinTuples sc x (fromMaybe [] val) key = getPartialTuple is x val = Map.lookup key m the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly because i know that it is not Nothing. is there a better way to solve this? regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
On Tue, Jun 2, 2009 at 4:59 PM, Nico Rolle nro...@web.de wrote: hi there heres a code snipped, don't care about the parameters. the thing is i make a lookup on my map m and then branch on that return value probePhase is sc [] m = [] probePhase is sc (x:xs) m | val == Nothing = probePhase is sc xs m | otherwise = jr ++ probePhase is sc xs m where jr = joinTuples sc x (fromMaybe [] val) key = getPartialTuple is x val = Map.lookup key m Here's my take. This ought to be equivalent, but I haven't tested. probePhase is sc m = concatMap prefix where prefix x = let key = getPartialTuple is x in maybe [] (joinTuples sc x) $ Map.lookup key m the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly because i know that it is not Nothing. is there a better way to solve this? regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
I just noticed that my suggestion doesn't work. You're testing whether val is Nothing and in my code snipped val has a different type. On 6/3/09, Raynor Vliegendhart shinnon...@gmail.com wrote: If you're absolutely certain that the lookup always succeeds, then you can use pattern matching as follows: where jr = joinTuples sc x val key = getPartialTuple is x Just val = Map.lookup key m On 6/3/09, Nico Rolle nro...@web.de wrote: hi there heres a code snipped, don't care about the parameters. the thing is i make a lookup on my map m and then branch on that return value probePhase is sc [] m = [] probePhase is sc (x:xs) m | val == Nothing = probePhase is sc xs m | otherwise = jr ++ probePhase is sc xs m where jr = joinTuples sc x (fromMaybe [] val) key = getPartialTuple is x val = Map.lookup key m the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly because i know that it is not Nothing. is there a better way to solve this? regards ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
On Wed, Jun 3, 2009 at 8:59 AM, Nico Rolle nro...@web.de wrote: hi there heres a code snipped, don't care about the parameters. the thing is i make a lookup on my map m and then branch on that return value probePhase is sc [] m = [] probePhase is sc (x:xs) m | val == Nothing = probePhase is sc xs m | otherwise = jr ++ probePhase is sc xs m where jr = joinTuples sc x (fromMaybe [] val) key = getPartialTuple is x val = Map.lookup key m the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly because i know that it is not Nothing. Although pattern matching is probably nicer, there's also fromJust which will throw an exception if you pass it Nothing. I prefer: case Map.lookup key m of Nothing - next Just val - (joinTuples sc x val) ++ next where next = probePhase ... key = ... ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Cabal/primes
Finally got adventurous enough to get Cabal working, downloaded the primes package, and got the following error message when trying isPrime. Am I missing something here? Michael == [mich...@localhost ~]$ ghci GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Prelude import Data.Numbers.Primes Prelude Data.Numbers.Primes take 10 primes Loading package syb ... linking ... done. Loading package base-3.0.3.0 ... linking ... done. Loading package primes-0.1.1 ... linking ... done. [2,3,5,7,11,13,17,19,23,29] Prelude Data.Numbers.Primes isPrime 7 interactive:1:0: Not in scope: `isPrime' Prelude Data.Numbers.Primes ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
where next = probePhase ... key = ... Argh, I really wish Gmail would allow me to compose in a fixed with width font! Does anyone know of a setting or something that I'm missing? ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Cabal/primes
michael rice wrote: Finally got adventurous enough to get Cabal working, downloaded the primes package, and got the following error message when trying isPrime. Am I missing something here? The Data.Numbers.Primes module of the primes package does not implement 'isPrime'. The Numbers package is probably the one you want. Bertram ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Checking a value against a passed-in constructor?
Ryan Ingram wrote: Dan danielkc...@gmail.com wrote: I figured there would be a clever Haskell idiom that would give me a similarly concise route. Does it really require Template Haskell? I can barely parse regular Haskell as it is.. [...] Alternatively, you can define a fold[1] once: myval :: MyVal - (Bool - a) - (String - a) - a myval (Bool b) bool atom = bool b myval (Atom s) bool atom = atom s f x = myval bool atom where bool b = ... atom s = ... In terms of boilerplate, this is often far and away the cleanest solution. I highly recommend if for Write Yourself A Scheme. The one place where it falls down is when, for whatever reason, you end up having collections of MyVals which can't sensibly use some set of constructors. One common example is for type-checking compilers where you guarantee that ill-typed MyVals cannot be constructed (rather than doing a verification pass after construction to ensure they're well-typed). If your type has this problem, using the fold approach often means writing dummy functions to throw errors on invalid inputs, which in turn means sacrificing much of the type safety you'd like (even if you use something like the Maybe or Error monads instead of _|_). A canonical Haskell trick here is to use GADTs to maintain your type invariants, rather than using plain ADTs. This technique isn't really suitable for a first pass at learning Haskell though. [1] fold here is the general term for this type of function. Examples are foldr: http://haskell.org/ghc/docs/latest/html/libraries/base/src/GHC-Base.html#foldr maybe: http://haskell.org/ghc/docs/latest/html/libraries/base/src/Data-Maybe.html#maybe either: http://haskell.org/ghc/docs/latest/html/libraries/base/src/Data-Either.html#either Two more good resources for folds are: * http://knol.google.com/k/edward-kmett/catamorphisms/ This has an implementation in Control.Morphism.Cata from category-extras[1], though the documentation is scarce. If you're interested in the theory of why folds look and work the way they do, then this knol is the best starting point. If you're familiar with OOP, a catamorphism is extremely similar to the recursive Visitor pattern. The big difference you'll see between this generic solution and specialized catamorphisms (foldr, maybe, either,...) is that the specialized versions unpack the Algebra into separate arguments. Also, this generic solution defines MyVal types with open-recursive functors and explicit fixed-point operators, whereas the specialized versions just use Haskell's regular ability to define recursive types (since the result of |fmap (cata f)| is consumed immediately). Don't let these trees obscure the forest. * http://www.cs.nott.ac.uk/~wss/Publications/DataTypesALaCarte.pdf This paper presents an innovative solution to the expression problem of defining an open set of constructors for a type. It uses the same open-recursive functor trick as above and may provide some illustration of why we may want to bother with it. If you're hungry for more details, there's an interesting discussion of the paper at [2]. [1] http://hackage.haskell.org/packages/archive/category-extras/ [2] http://wadler.blogspot.com/2008/02/data-types-la-carte.html -- Live well, ~wren ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] beginners question about fromMaybe
Luke's answer is great (although it changes argument order). Hint:
http://www.haskell.org/haskellwiki/Things_to_avoid#Avoid_explicit_recursion
I also like the pattern guards GHC extension; I tend to use it over
maybe and either. I find the resulting code more readable:
{-# LANGUAGE PatternGuards #-}
probePhase is sc xs m = concatMap prefix xs where
prefix x
| Just val - Map.lookup (getPartialTuple is x) m = joinTuples sc x
val
| otherwise = []
Alternatively, I might write it like this:
import Control.Monad
maybeM :: MonadPlus m = Maybe a - m a
maybeM = maybe mzero return
probePhase is sc xs m = do
x - xs
val - maybeM $ Map.lookup (getPartialTuple is x) m
joinTuples sc x val
This now works for any xs that is an instance of MonadPlus (assuming
joinTuples is also polymorphic).
Both of these examples are more wordy than Luke's quick two-liner,
but, to me, it's worth it for the additional maintainability of that
code. I am perhaps in the minority on this issue, though :)
-- ryan
On Tue, Jun 2, 2009 at 4:20 PM, Luke Palmer lrpal...@gmail.com wrote:
On Tue, Jun 2, 2009 at 4:59 PM, Nico Rolle nro...@web.de wrote:
hi there
heres a code snipped, don't care about the parameters.
the thing is i make a lookup on my map m and then branch on that return
value
probePhase is sc [] m = []
probePhase is sc (x:xs) m
| val == Nothing = probePhase is sc xs m
| otherwise = jr ++ probePhase is sc xs m
where
jr = joinTuples sc x (fromMaybe [] val)
key = getPartialTuple is x
val = Map.lookup key m
Here's my take. This ought to be equivalent, but I haven't tested.
probePhase is sc m = concatMap prefix
where
prefix x = let key = getPartialTuple is x in
maybe [] (joinTuples sc x) $ Map.lookup key m
the line jr = joinTuples sc x (fromMaybe [] val) is kind of ugly
because i know that it is not Nothing.
is there a better way to solve this?
regards
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[Haskell-cafe] type checking that I can't figure out ....
Hello Haskellers,
I isolated to a not so small piece:
{-# OPTIONS -fglasgow-exts #-}
{-# LANGUAGE UndecidableInstances #-}
import Control.Monad.Identity
import Control.Monad.Reader
import Control.Monad.State
import qualified Data.List as L
import qualified Data.Map as M
import Data.Array
import IOExts
The type of a regular expression.
data Re t
= ReOr [Re t]
| ReCat [Re t]
| ReStar (Re t)
| RePlus (Re t)
| ReOpt (Re t)
| ReTerm [t]
deriving (Show)
The internal type of a regular expression.
type SimplRe t = Int
data SimplRe' t
= SReOr (SimplRe t) (SimplRe t)
| SReCat (SimplRe t) (SimplRe t)
| SReStar (SimplRe t)
| SReLambda
| SReNullSet
| SReTerm t
deriving (Eq, Ord, Show)
The regular expression builder monad.
data (Ord t) = ReRead t
= ReRead {
rerNullSet :: SimplRe t,
rerLambda :: SimplRe t
}
data (Ord t) = ReState t
= ReState {
resFwdMap :: M.Map (SimplRe t) (ReInfo t),
resBwdMap :: M.Map (SimplRe' t) (SimplRe t),
resNext :: Int,
resQueue:: ([SimplRe t], [SimplRe t]),
resStatesDone :: [SimplRe t]
}
type ReM m t a = StateT (ReState t) (ReaderT (ReRead t) m) a
TEMP WNH
Dfa construction
data ReDfaState t
= ReDfaState {
dfaFinal :: Bool,
dfaTrans :: [(t, SimplRe t)]
}
deriving (Show)
TEMP WNH
The ReInfo type
data ReInfo t
= ReInfo {
reiSRE :: SimplRe' t,
reiNullable :: Bool,
reiDfa :: Maybe (ReDfaState t)
}
deriving (Show)
TEMP WNH
class (Monad m, Ord t) = ReVars m t where { }
instance (Monad m, Ord t) = ReVars m t where { }
TEMP WNH
remLookupFwd :: (ReVars m t) = SimplRe t - ReM m t (ReInfo t)
remLookupFwd re
= do fwd - gets resFwdMap
-- let { Just reinfo = M.lookup fwd re }--
PROBLEM
reinfo - M.lookup fwd re -- PROBLEM
return reinfo
When I compile with ghci I get:
Dfa_exp.lhs:91:32:
Couldn't match expected type `M.Map
(M.Map (SimplRe t) (ReInfo t)) t1'
against inferred type `SimplRe t2'
In the second argument of `M.lookup', namely `re'
In a 'do' expression: reinfo - M.lookup fwd re
In the expression:
do fwd - gets resFwdMap
reinfo - M.lookup fwd re
return reinfo
I trimmed the original code down a lot! But still can't why I am getting
type check errors!!! Help!
Kind regards,
Vasili
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Re: [Haskell-cafe] type checking that I can't figure out ....
remLookupFwd :: (ReVars m t) = SimplRe t - ReM m t (ReInfo t)
remLookupFwd re
= do fwd - gets resFwdMap
-- let { Just reinfo = M.lookup fwd re }--
PROBLEM
reinfo - liftMaybe $ M.lookup re fwd --
PROBLEM
return reinfo
liftMaybe :: Monad m = Maybe a - m a
liftMaybe Nothing = fail Nothing
liftMaybe (Just x) = return x
I made two changes:
1. You had the arguments to M.lookup backwards.
2. lookup does not return any generalized Monad, just Maybe (I think that
should be changed). I added the simple liftMaybe function to convert the
Maybe result into something that will work with your state monad.
Michael
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Re: [Haskell-cafe] type checking that I can't figure out ....
Hi Michael,
Let me look tomorrow morning. In any case, many thanks!
Kind regards,
Vasili
On Tue, Jun 2, 2009 at 11:12 PM, Michael Snoyman mich...@snoyman.comwrote:
remLookupFwd :: (ReVars m t) = SimplRe t - ReM m t (ReInfo t)
remLookupFwd re
= do fwd - gets resFwdMap
-- let { Just reinfo = M.lookup fwd re }--
PROBLEM
reinfo - liftMaybe $ M.lookup re fwd --
PROBLEM
return reinfo
liftMaybe :: Monad m = Maybe a - m a
liftMaybe Nothing = fail Nothing
liftMaybe (Just x) = return x
I made two changes:
1. You had the arguments to M.lookup backwards.
2. lookup does not return any generalized Monad, just Maybe (I think that
should be changed). I added the simple liftMaybe function to convert the
Maybe result into something that will work with your state monad.
Michael
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Re: [Haskell-cafe] type checking that I can't figure out ....
G'day Vasili.
This should do it:
remLookupFwd :: (ReVars m t) = SimplRe t - ReM m t (ReInfo t)
remLookupFwd re
= do fwd - gets resFwdMap
let { Just reinfo = fromJust (M.lookup re fwd) }
return reinfo
The FiniteMap lookup operation took its arguments in the opposite order.
That's really the only problem here AFAICT.
Wow, this brings back memories. I wrote this module about ten years ago,
and I'm shocked that it's still getting use. I'd appreciate a copy when
you're done updating it for the modern era.
Cheers,
Andrew Bromage
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Re: [Haskell-cafe] ANN: Anglohaskell 2009
Henk-Jan van Tuyl wrote: On Tue, 02 Jun 2009 23:45:18 +0200, Philippa Cowderoy fli...@flippac.org wrote: Anglohaskell 2009 is go! F.A.B. :) Yes, excellent news, and this time I'll make sure to attend, especially since it's back in Cambridge again. /M -- Magnus Therning(OpenPGP: 0xAB4DFBA4) magnus＠therning．org Jabber: magnus＠therning．org http://therning.org/magnus identi.ca|twitter: magthe signature.asc Description: OpenPGP digital signature ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] type checking that I can't figure out ....
Am Mittwoch 03 Juni 2009 06:12:46 schrieb Michael Snoyman: I made two changes: 1. You had the arguments to M.lookup backwards. 2. lookup does not return any generalized Monad, just Maybe (I think that should be changed). Data.Map.lookup used to return a value in any monad you wanted, I believe until 6.8 inclusive. I don't think it's going to change again soon. I added the simple liftMaybe function to convert the Maybe result into something that will work with your state monad. Michael ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
