[antlr-interest] assigning token LABEL

[chintan rao]

Hi,
firstly thanks for your time

>As you haven't sent your grammar with your question I have to improvise
>the snippets. Create first an imaginary LABEL token and have rules like
>this:

What wanted was some thing like this
input :
la :
ouput given by lexer :
la gets assigned token LABEL

input:
la
ouput given by lexer :
la gets assigned token ID

I tried out your grammar with few changes in antlrworks
<code starts >

grammar label_test;


LABEL	//had to make this LABEL instead of 'label'
  :   ID[false] ':' ID[true]
  ;

fragment ID [ boolean isID ]  // gave errors without fragment
  :   ( ('a'..'z'|'A'..'Z')+ ){ if (!isID) {$type=LABEL ;} }
  ;


test	:	LABEL;

<code ends>


Now this gives me the following error
{
label_testLexer.java:90: cannot find symbol
 symbol  : variable _type
 location: class label_testLexer
             if (!isID) {_type=LABEL ;}
                         ^
1 error
}

I tried to make more changes to the code and make it work but could not
Please help me out


On 6/20/07, Johannes Luber <jaluber at gmx.de> wrote:
> chintan rao wrote:
> > Hi,
> > I am new to antlr3 / compilers / parsers .
> > I have problem assigning token LABEL .
> > if there is ':' after ID it must be labeled as LABEL .
> > if there is no colon then it must be named as ID that is  IDENTIFIER .
> >
> > This may be a simple problem but i am stuck with this .
> > please help me out.
>
> As you haven't sent your grammar with your question I have to improvise
> the snippets. Create first an imaginary LABEL token and have rules like
> this:
>
> label
>    :   ID[false] ':' ID[true]
>    ;
>
> ID[boolean isID]
>    :   ('a'..'z'|'A'..'Z')+ { if (!isID) $type = LABEL ; }
>    ;
>
> Hope this helps!
> Johannes Luber
>
>