Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Am Montag, 21. November 2005 03:27 schrieb Sara Kenedy:
May I suggest
endBy anyToken semi ? -- optionally replace semi by char ';', if you don't
want to skip whitespace
I think this is what you want --- stop at the first semicolon.
If you want to ignore just a final semicolon, you might use
endBy anyToken (optional semi eof),
if you want to stop at the last semicolon, whatever comes thereafter, you have
a problem, you'd need long lookahead.
Cheers,
Daniel
Thanks for your solution. However, when I try this,
str1 :: Parser String
str1 = do str - many anyToken
notFollowedBy' semi
return str
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return
(unexpected (show a)) |
return (return ())
run:: Show a = Parser a - String - IO()
run p input
= case (parse p input) of
Left err - do {putStr parse error at ;print err}
Right x - print
When I compile, it still displays ; at the end of the string.
Parser run str1 Hello ;
Hello ;
The reason, as I think, because anyToken accepts any kind of token, it
considers ; as token of its string. Thus, it does not understand
notFollowedBy' ???
Do you have any ideas about this ??? Thanks.
On 11/19/05, Andrew Pimlott [EMAIL PROTECTED] wrote:
On Sat, Nov 19, 2005 at 06:43:48PM -0500, Sara Kenedy wrote:
str1 :: Parser String
str1 = do {str - many anyToken; notFollowedBy semi; return str}
However, when I compile, there is an error.
ERROR Test.hs:17 - Type error in application
*** Expression : notFollowedBy semi
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser [Char] () [Char]
The problem is that notFollowedBy has type
notFollowedBy :: Show tok = GenParser tok st tok - GenParser tok
st ()
ie, the result type of the parser you pass to notFollowedBy has to be
the same as the token type, in this case Char. (The reason for this
type is obscure.) But semi has result type String. You could fix the
type error by returning a dummy Char:
str1 = do {str - many anyToken
; notFollowedBy (semi return undefined)
; return str}
I think this will even work; however notFollowedBy is a pretty
squirrelly function. There was a discussion about it:
http://www.haskell.org/pipermail/haskell/2004-February/013621.html
Here is a version (which came out of that thread) with a nicer type,
that probably also works more reliably (though I won't guarantee it):
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return (unexpected (show a))
|
return (return ())
Andrew
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Am Dienstag, 22. November 2005 14:51 schrieben Sie:
Am Montag, 21. November 2005 03:27 schrieb Sara Kenedy:
May I suggest
endBy anyToken semi ? -- optionally replace semi by char ';', if you
Oops, I confused endBy and manyTill !! Also below.
And since maybe there isn't any semicolon, I'd say
manyTill anyToken (semi {- try semi, perhaps -} | eof)
don't want to skip whitespace
I think this is what you want --- stop at the first semicolon.
If you want to ignore just a final semicolon, you might use
endBy anyToken (optional semi eof),
if you want to stop at the last semicolon, whatever comes thereafter, you
have a problem, you'd need long lookahead.
Cheers again,
Daniel
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Hello,
I run as follows:
simple::Parser String
simple = do manyTill anyToken (semi | eof)
run:: Show a = Parser a - String - IO()
run p input
= case (parse p input) of
Left err - do {putStr parse error at ;print err}
Right x - print x
ParsecLanguage :load Test.hs
Type checking
ERROR Test.hs:21 - Type error in application
*** Expression : semi | eof
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser a b ()
Do you know what happens? Thank you.
On 11/22/05, Daniel Fischer [EMAIL PROTECTED] wrote:
Am Dienstag, 22. November 2005 14:51 schrieben Sie:
Am Montag, 21. November 2005 03:27 schrieb Sara Kenedy:
May I suggest
endBy anyToken semi ? -- optionally replace semi by char ';', if you
Oops, I confused endBy and manyTill !! Also below.
And since maybe there isn't any semicolon, I'd say
manyTill anyToken (semi {- try semi, perhaps -} | eof)
don't want to skip whitespace
I think this is what you want --- stop at the first semicolon.
If you want to ignore just a final semicolon, you might use
endBy anyToken (optional semi eof),
if you want to stop at the last semicolon, whatever comes thereafter, you
have a problem, you'd need long lookahead.
Cheers again,
Daniel
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Am Dienstag, 22. November 2005 15:58 schrieben Sie:
Hello,
I run as follows:
simple::Parser String
simple = do manyTill anyToken (semi | eof)
run:: Show a = Parser a - String - IO()
run p input
= case (parse p input) of
Left err - do {putStr parse error at ;print err}
Right x - print x
ParsecLanguage :load Test.hs
Type checking
ERROR Test.hs:21 - Type error in application
*** Expression : semi | eof
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser a b ()
Do you know what happens? Thank you.
Aye, | takes two parsers of the same type, so we'd need
manyTill anyToken ((semi return () ) | eof)
or
manyTill anyToken (semi | (eof return dummy String))
Cheers,
Daniel
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Sara Kenedy wrote:
import qualified ParsecToken as P
the proper hierarchical module name is:
Text.ParserCombinators.Parsec.Token
str1 :: Parser String
str1 = do {str - many anyToken; notFollowedBy semi; return str}
simply try:
str - many anyToken; notFollowedBy (char ';'); return str
semi only skips additional white spaces (that you are not interested in)
(I find it easier not to use the Parsec.Token und Parsec.Language
wrappers and remain Haskell 98 conform)
Christian
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
On Sun, Nov 20, 2005 at 09:27:53PM -0500, Sara Kenedy wrote:
Thanks for your solution. However, when I try this,
str1 :: Parser String
str1 = do str - many anyToken
notFollowedBy' semi
return str
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return (unexpected
(show a))
|
return (return ())
run:: Show a = Parser a - String - IO()
run p input
= case (parse p input) of
Left err - do {putStr parse error at ;print err}
Right x - print
When I compile, it still displays ; at the end of the string.
Parser run str1 Hello ;
Hello ;
The reason, as I think, because anyToken accepts any kind of token, it
considers ; as token of its string. Thus, it does not understand
notFollowedBy' ???
That's right--your parser consumes and returns the whole input. I can't
tell you what to use instead, because it depends on what kinds of
strings you want to parse. Since you are using Token parsers, maybe you
want symbol? The functions in the Char module might also be useful.
Andrew
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
Thanks for your solution. However, when I try this,
str1 :: Parser String
str1 = do str - many anyToken
notFollowedBy' semi
return str
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return (unexpected
(show a))
|
return (return ())
run:: Show a = Parser a - String - IO()
run p input
= case (parse p input) of
Left err - do {putStr parse error at ;print err}
Right x - print
When I compile, it still displays ; at the end of the string.
Parser run str1 Hello ;
Hello ;
The reason, as I think, because anyToken accepts any kind of token, it
considers ; as token of its string. Thus, it does not understand
notFollowedBy' ???
Do you have any ideas about this ??? Thanks.
On 11/19/05, Andrew Pimlott [EMAIL PROTECTED] wrote:
On Sat, Nov 19, 2005 at 06:43:48PM -0500, Sara Kenedy wrote:
str1 :: Parser String
str1 = do {str - many anyToken; notFollowedBy semi; return str}
However, when I compile, there is an error.
ERROR Test.hs:17 - Type error in application
*** Expression : notFollowedBy semi
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser [Char] () [Char]
The problem is that notFollowedBy has type
notFollowedBy :: Show tok = GenParser tok st tok - GenParser tok st ()
ie, the result type of the parser you pass to notFollowedBy has to be
the same as the token type, in this case Char. (The reason for this
type is obscure.) But semi has result type String. You could fix the
type error by returning a dummy Char:
str1 = do {str - many anyToken
; notFollowedBy (semi return undefined)
; return str}
I think this will even work; however notFollowedBy is a pretty
squirrelly function. There was a discussion about it:
http://www.haskell.org/pipermail/haskell/2004-February/013621.html
Here is a version (which came out of that thread) with a nicer type,
that probably also works more reliably (though I won't guarantee it):
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return (unexpected (show a))
|
return (return ())
Andrew
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[Haskell-cafe] How to use notFollowedBy function in Parsec
Dear all,
Using Parsec, I want to represent a string (of anyToken) not ended
with symbol semi (;). I use the command notFollowedby as follows:
module Parser where
import Parsec
import qualified ParsecToken as P
import ParsecLanguage
langDef::LanguageDef ()
langDef = emptyDef
{reservedOpNames = []}
lexer::P.TokenParser()
lexer = P.makeTokenParser langDef
semi= P.semi lexer
str1 :: Parser String
str1 = do {str - many anyToken; notFollowedBy semi; return str}
However, when I compile, there is an error.
ERROR Test.hs:17 - Type error in application
*** Expression : notFollowedBy semi
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser [Char] () [Char]
I do not know how to fix it. Help me. Thanks for your time.
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Re: [Haskell-cafe] How to use notFollowedBy function in Parsec
On Sat, Nov 19, 2005 at 06:43:48PM -0500, Sara Kenedy wrote:
str1 :: Parser String
str1 = do {str - many anyToken; notFollowedBy semi; return str}
However, when I compile, there is an error.
ERROR Test.hs:17 - Type error in application
*** Expression : notFollowedBy semi
*** Term : semi
*** Type : GenParser Char () String
*** Does not match : GenParser [Char] () [Char]
The problem is that notFollowedBy has type
notFollowedBy :: Show tok = GenParser tok st tok - GenParser tok st ()
ie, the result type of the parser you pass to notFollowedBy has to be
the same as the token type, in this case Char. (The reason for this
type is obscure.) But semi has result type String. You could fix the
type error by returning a dummy Char:
str1 = do {str - many anyToken
; notFollowedBy (semi return undefined)
; return str}
I think this will even work; however notFollowedBy is a pretty
squirrelly function. There was a discussion about it:
http://www.haskell.org/pipermail/haskell/2004-February/013621.html
Here is a version (which came out of that thread) with a nicer type,
that probably also works more reliably (though I won't guarantee it):
notFollowedBy' :: Show a = GenParser tok st a - GenParser tok st ()
notFollowedBy' p = try $ join $ do a - try p
return (unexpected (show a))
|
return (return ())
Andrew
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