'expression' is rule 'r' wont return a Token since it is not a lexer rule. It
will instead return 'ParserRuleReturnScope' since it is a parser rule. So
'$expression.stop' or '$t.stop' will give you the last token of 'expr'. You can
use it like shown below:
r : t=expression { newLineBeforeNextToken( $t.stop.getTokenIndex() )
}?=>/*nothing*/
;
or
r : expression { newLineBeforeNextToken( $expression.stop.getTokenIndex() )
}?=>/*nothing*/
;
Cheers, Indhu
----- Original Message -----
From: Andreas Stefik <[email protected]>
To: Michael Bedward <[email protected]>, [email protected]
Sent: Saturday, April 4, 2009 10:33:02 PM GMT+0530 Asia/Calcutta
Subject: Re: [antlr-interest] A Simple Question on Channels
Indhu,
Thanks, this is very helpful. One more very related question, if you'll indulge
me. The grammar you posted works great, and I think I'm close to understanding
what's going on here. You define @members, which dumps a custom function into
the generated parser. Inside, you get the tokens from any channel that is not
the default, starting from the index of the token in the parser rule.
However, in my grammar, t cannot be set to a token, because I need that newline
to be processed after an expression, which can have any arbitrary number of
tokens in it, so my code (modifying your grammar), would need to look something
like this:
r : t=expression { newLineBeforeNextToken( $t.getTokenIndex() ) }?=>/*nothing*/
;
expression : equality
;
equality: add (('==' | '!=') add)*
;
add : multiply (('+' | '-') multiply)*
;
multiply: atom (('*' | '/') atom)*
;
atom : ID
| '(' expression ')'
;
I know that first rule isn't right, but I need to have ANTLR determine (if
possible), the index of the last token in the expression, and have that value
be passed to your function. How do I modify the above to accommodate for having
an arbitrary rule in place of where you had ID?
Thanks again, this was very helpful, and gets me started,
Andreas
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