Re: [Haskell-cafe] Bencoding in Haskell
Yeah, you probably want the main parser to be many beParser and not just beParser: -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencode str = case parse (many beParse) str of Left err - Nothing Right val - Just val On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: On 4/20/05, Tommi Airikka [EMAIL PROTECTED] wrote: Hi! I was just wondering if there are any good ways to represent a bencoded (http://en.wikipedia.org/wiki/Bencoding) message in Haskell? Any suggestions? Not that I know of, but it should be very easy to write a parser using the parser library Parsec. You'll need a datatype, something like this: data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Data.Map String Bencode) deriving (Show, Eq) Which should be sufficient to represent any Bencoded message (if I didn't make a misstake). Then you could probably use the standard char-parser in parsec to parse it quite easily. Read the docs, they're quite straightforward. I'm a bit rusty but something like this: -- just parse an integer, parsec might have one of these already number :: Parser Integer number = do n_str - many1 digit -- parse a number let n = read n_str -- convert to an Int return n -- return the number beString :: Parser Bencode beString = do n - number -- the length prefix char ':' -- now a ':' str - count n anyChar -- and now n number of letters return (BEString str) -- return the string wrapped up as a BEString beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return n -- parse any Bencoded value beParse :: Parser Bencode beParse = do beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' key - beString val - beParse m - beDictionary | char 'e' return Data.Map.empty return (Data.Map.insert key val m) -- main parser function parseBencoded :: String - Maybe Bencode parseBencode str = case parse beParse str of Left err - Nothing Right val - Just val Note: This is all untested code that I just scribbled down real quick. There's probably tons of misstakes, but you should get the picture. Read the Parsec docs and then write your own. /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Bencoding in Haskell
I was bored so I ran it through ghci and fixed the small errors I found, here's the working version, I don't really have much of test data to play with, but it seems to be working with the small examples I copy-n-pasted from the wiki and the bittorrent website: import qualified Data.Map as Map import Data.Map(Map) import Text.ParserCombinators.Parsec data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Map String Bencode) deriving (Show, Eq) number :: Parser Integer number = do n_str - many1 digit let n = read n_str return n beString :: Parser Bencode beString = do n - number char ':' str - count (fromInteger n) anyChar return (BEString str) beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return (BEInteger n) -- parse any Bencoded value beParse :: Parser Bencode beParse = beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' (BEString key) - beString val - beParse (BEDictionary m) - beDictionary | do char 'e' return (BEDictionary Map.empty) return (BEDictionary (Map.insert key val m)) -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencoded str = case parse (many beParse) str of Left err - Nothing Right val - Just val /S On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: Yeah, you probably want the main parser to be many beParser and not just beParser: -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencode str = case parse (many beParse) str of Left err - Nothing Right val - Just val On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: On 4/20/05, Tommi Airikka [EMAIL PROTECTED] wrote: Hi! I was just wondering if there are any good ways to represent a bencoded (http://en.wikipedia.org/wiki/Bencoding) message in Haskell? Any suggestions? Not that I know of, but it should be very easy to write a parser using the parser library Parsec. You'll need a datatype, something like this: data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Data.Map String Bencode) deriving (Show, Eq) Which should be sufficient to represent any Bencoded message (if I didn't make a misstake). Then you could probably use the standard char-parser in parsec to parse it quite easily. Read the docs, they're quite straightforward. I'm a bit rusty but something like this: -- just parse an integer, parsec might have one of these already number :: Parser Integer number = do n_str - many1 digit -- parse a number let n = read n_str -- convert to an Int return n -- return the number beString :: Parser Bencode beString = do n - number -- the length prefix char ':' -- now a ':' str - count n anyChar -- and now n number of letters return (BEString str) -- return the string wrapped up as a BEString beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return n -- parse any Bencoded value beParse :: Parser Bencode beParse = do beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' key - beString val - beParse m - beDictionary | char 'e' return Data.Map.empty return (Data.Map.insert key val m) -- main parser function parseBencoded :: String - Maybe Bencode parseBencode str = case parse beParse str of Left err - Nothing Right val - Just val Note: This is all untested code that I just scribbled down real quick. There's probably tons of misstakes, but you should get the picture. Read the Parsec docs and then write your own. /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Bencoding in Haskell
Thank you very much! I really appreciate your help! I have to read a little bit more about Parsec to fully understand what your code does, but it seems to be what I was looking for. Regards, Tommi On Wed, Apr 20, 2005 at 08:58:41PM +0200, Sebastian Sylvan wrote: I was bored so I ran it through ghci and fixed the small errors I found, here's the working version, I don't really have much of test data to play with, but it seems to be working with the small examples I copy-n-pasted from the wiki and the bittorrent website: import qualified Data.Map as Map import Data.Map(Map) import Text.ParserCombinators.Parsec data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Map String Bencode) deriving (Show, Eq) number :: Parser Integer number = do n_str - many1 digit let n = read n_str return n beString :: Parser Bencode beString = do n - number char ':' str - count (fromInteger n) anyChar return (BEString str) beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return (BEInteger n) -- parse any Bencoded value beParse :: Parser Bencode beParse = beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' (BEString key) - beString val - beParse (BEDictionary m) - beDictionary | do char 'e' return (BEDictionary Map.empty) return (BEDictionary (Map.insert key val m)) -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencoded str = case parse (many beParse) str of Left err - Nothing Right val - Just val /S On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: Yeah, you probably want the main parser to be many beParser and not just beParser: -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencode str = case parse (many beParse) str of Left err - Nothing Right val - Just val On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: On 4/20/05, Tommi Airikka [EMAIL PROTECTED] wrote: Hi! I was just wondering if there are any good ways to represent a bencoded (http://en.wikipedia.org/wiki/Bencoding) message in Haskell? Any suggestions? Not that I know of, but it should be very easy to write a parser using the parser library Parsec. You'll need a datatype, something like this: data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Data.Map String Bencode) deriving (Show, Eq) Which should be sufficient to represent any Bencoded message (if I didn't make a misstake). Then you could probably use the standard char-parser in parsec to parse it quite easily. Read the docs, they're quite straightforward. I'm a bit rusty but something like this: -- just parse an integer, parsec might have one of these already number :: Parser Integer number = do n_str - many1 digit -- parse a number let n = read n_str -- convert to an Int return n -- return the number beString :: Parser Bencode beString = do n - number -- the length prefix char ':' -- now a ':' str - count n anyChar -- and now n number of letters return (BEString str) -- return the string wrapped up as a BEString beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return n -- parse any Bencoded value beParse :: Parser Bencode beParse = do beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' key - beString val - beParse m - beDictionary | char 'e' return Data.Map.empty return (Data.Map.insert key val m) -- main parser function parseBencoded :: String - Maybe Bencode parseBencode str = case parse beParse str of Left err - Nothing Right val - Just val Note: This is all untested code that I just scribbled down real quick. There's probably tons of misstakes, but you should get the picture.
[Haskell-cafe] Haskell Assistance Question
Hi, Im currently working on a haskell constraint parser for polymorhpic data types for a group of third year programmers, and was wondering if there was a help resource, or board on the internet for haskell programmers to ask questions about code they're working on. Any help or links would be much appreciated. Thanks a lot Mike [EMAIL PROTECTED] -- ___ Get your free email from http://mail.oxygen.ie Powered by Outblaze ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] Bencoding in Haskell
beDictionary is wrong, though. It will only find dictionaries with a single entry. This next parser should do the trick (again, untested!). It basically reads a d and then a list of (key,value) pairs (which is now a separate parser) and then an e, and then it returns a Map String Bencode. Should be something like this -- parses an association list of the contents -- of a dictionary beDicContents :: Parser (String, Bencode) beDicContents = do (BEString key) - beString val - beParse return (String, Bencode) beDictionary :: Parser Bencode beDictionary = do char 'd' xs - many beDicContents char 'e' return (BEDictionary (Map.fromAscList xs)) On 4/20/05, Tommi Airikka [EMAIL PROTECTED] wrote: Thank you very much! I really appreciate your help! I have to read a little bit more about Parsec to fully understand what your code does, but it seems to be what I was looking for. Regards, Tommi On Wed, Apr 20, 2005 at 08:58:41PM +0200, Sebastian Sylvan wrote: I was bored so I ran it through ghci and fixed the small errors I found, here's the working version, I don't really have much of test data to play with, but it seems to be working with the small examples I copy-n-pasted from the wiki and the bittorrent website: import qualified Data.Map as Map import Data.Map(Map) import Text.ParserCombinators.Parsec data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Map String Bencode) deriving (Show, Eq) number :: Parser Integer number = do n_str - many1 digit let n = read n_str return n beString :: Parser Bencode beString = do n - number char ':' str - count (fromInteger n) anyChar return (BEString str) beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return (BEInteger n) -- parse any Bencoded value beParse :: Parser Bencode beParse = beInt | beString | beDictionary | beList beList :: Parser Bencode beList = do char 'l' xs - many beParse -- parse many bencoded values char 'e' return (BEList xs) beDictionary :: Parser Bencode beDictionary = do char 'd' (BEString key) - beString val - beParse (BEDictionary m) - beDictionary | do char 'e' return (BEDictionary Map.empty) return (BEDictionary (Map.insert key val m)) -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencoded str = case parse (many beParse) str of Left err - Nothing Right val - Just val /S On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: Yeah, you probably want the main parser to be many beParser and not just beParser: -- main parser function parseBencoded :: String - Maybe [Bencode] parseBencode str = case parse (many beParse) str of Left err - Nothing Right val - Just val On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: On 4/20/05, Tommi Airikka [EMAIL PROTECTED] wrote: Hi! I was just wondering if there are any good ways to represent a bencoded (http://en.wikipedia.org/wiki/Bencoding) message in Haskell? Any suggestions? Not that I know of, but it should be very easy to write a parser using the parser library Parsec. You'll need a datatype, something like this: data Bencode = BEInteger Integer | BEString String | BEList [Bencode] | BEDictionary (Data.Map String Bencode) deriving (Show, Eq) Which should be sufficient to represent any Bencoded message (if I didn't make a misstake). Then you could probably use the standard char-parser in parsec to parse it quite easily. Read the docs, they're quite straightforward. I'm a bit rusty but something like this: -- just parse an integer, parsec might have one of these already number :: Parser Integer number = do n_str - many1 digit -- parse a number let n = read n_str -- convert to an Int return n -- return the number beString :: Parser Bencode beString = do n - number -- the length prefix char ':' -- now a ':' str - count n anyChar -- and now n number of letters return (BEString str) -- return the string wrapped up as a BEString beInt :: Parser Bencode beInt = do char 'i' n - number char 'e' return n -- parse any Bencoded value
Re: [Haskell-cafe] Bencoding in Haskell
On 4/20/05, Sebastian Sylvan [EMAIL PROTECTED] wrote: -- parses an association list of the contents -- of a dictionary beDicContents :: Parser (String, Bencode) beDicContents = do (BEString key) - beString val - beParse return (String, Bencode) bah, I should get some sleep :-) That should be return (key, val) /S -- Sebastian Sylvan +46(0)736-818655 UIN: 44640862 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: [Haskell] Going nuts
Alexandre Weffort Thenorio [EMAIL PROTECTED] writes:
outputLine keyno key orgFile = do
part1 - getLeft keyno orgFile
part2 - getRight keyno orgFile
total - part1 ++ (strUpper key) ++ part2 ++ \n
newHexFile - openFileEx newfile (BinaryMode WriteMode)
hPutStrLn newHexFile (orgFile!!0 ++ \n ++ total ++ unlines (drop 2
orgFile))
outputLine keyno key orgFile = do
let part1 = getLeft keyno orgFile
let part2 = getRight keyno orgFile
let total = part1 ++ (strUpper key) ++ part2 ++ \n
newHexFile - openFileEx newfile (BinaryMode WriteMode)
hPutStrLn newHexFile (orgFile!!0 ++ \n ++ total ++ unlines (drop 2
orgFile))
let ... = instead of - because getLeft et al. aren't IO commands.
(Why type error rather than syntax error then? Because getLeft returns
a list, and list is a monad too. E.g.,
do { part1 - getLeft keyno orgFile; return part1 }
= [ part1 | part1 - getLeft keyno orgFile ]
so part1 is inferred to be a Char.)
There will be another problem. The type of orgFile is expected to be
String here, but the callsite gives it lines(hexFile) of type [String].
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[Haskell-cafe] embedded interpreters
[I've asked this question on lambda-the-ultimate, but it remained unanswered, and it's probably more appropriate here in any case] The papers referenced here: http://lambda-the-ultimate.org/node/view/552 describe a means of projecting values in an implementation language to and from values in an embedded interpreted language. The papers use ML, but Derek Elkin gives a straightforward translation to haskell, via a typeclass. An excerpt is here: data U = UF !(U - U) | UP !U !U | UU | UB !Bool | UI !Int | US !String deriving Show class EmbedProject a where embed :: a - U project :: U - a instance EmbedProject Int where embed = UI project (UI i) = i instance (EmbedProject a, EmbedProject b) = EmbedProject (a - b) where embed f = UF (embed . f . project) project (UF f) = project . f . embed This is nice and tidy. However, the project function can fail in practise - in the above, this results in pattern matching failures. I'd like have more control of this, which suggests the following: class EmbedProject a where embed :: a - U project :: U - Maybe a instance EmbedProject Int where embed = UI project (UI i) = Just i project _ = Nothing ...etc, until it comes to to defining the instance of EmbedProject for functions. embed is ok - incorrect parameters map to return values of the UU type. But I'm lost as to what needs to happen with the project function. The definition belows doesn't compile because (project.f.embed) has type a-Maybe b, but I need something of type Maybe (a-b): instance (EmbedProject a, EmbedProject b) = EmbedProject (a - b) where embed f = UF (\a - maybe UU (embed.f) (project a) ) project (UF f) = project.f.embed project _ = Nothing Any suggestions? Is there something obvious (or non-obvious!) that I am missing? Thanks, Tim ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] Re: [Haskell] Y in haskell?
Bernard Pope [EMAIL PROTECTED] writes: I also meant to add that I think these solutions are not what Lloyd is after, because they rely on recursive equations, which I believe was avoided in Lloyd's SML code. Those recursive equations are avoided in SML because SML is eager - y f = f (y f) never terminates - and an SML programmer has to thunk by hand; the obfuscation shows. In Haskell there is no such need. ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
