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

[Hans Aberg]

At 15:38 +0200 2002/07/04, address@hidden wrote:
>I run bison -v "myfile.y" and got output where last two rules I can't
>understand:
>state 15
>    $           go to state 16

>state 16
>    $           go to state 17

>state 17
>
>    $default    accept

huh what $ is for? Is
>it some kind of "end of input"?

Yes, the parser algorithm requires a token end-marker which is called $.

>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.

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>Waiting for reply...

Enjoy!

  Hans Aberg