Re: Bison and $ (dollarsign) token at the end

[Grzegorz_Szostak]

>>If parser is in state 15 it needs two $ to accept...
>
>Not really: The $ token lies in the input lookahead unaltered. For other
>tokens there are two operation "shift" = put onto stack, and "reduce" =
>apply a rule to top of stack. For $ it only says "go to state n", so
>nothing happens with it.
>
>>How can I make yylex (flex++) make returns twice (-1)?
>
>So this is not needed. Otherwise, the Bison parser is accepting values <=
0
>as end token.

OK, so why I get message: parse error, expecting `$' ?
My rules comes here:
expression
:       {}
|       lines
;

lines
:       line
|       lines line
;

line
:       def_decl
|       assig
|       val
|       operation
;

def_decl
:       VAR VAR_NAME EQU NUMBER {
        (*pVars)[(*$1)] + (*pVars)[(*$3)];
        $$ = &((*pVars)[(*$1)].getString());
}
;
assig
:       VAR_NAME EQU NUMBER
operation
:       VAR_NAME PLUS VAR_NAME
;
val
:       VAR_NAME
;


When I make input succefying def_decl, parser puts " " - white space on
output...
Why?

Regardless
GSZ