Re: how to abort when token are discarded

[Bill Allombert]

On Sat, Feb 19, 2011 at 12:26:47PM -0500, Joel E. Denny wrote:
> On Sat, 12 Feb 2011, Bill Allombert wrote:
> 
> > I have a grammar that handles errors.
> 
> Do you mean you're using Bison's error token?  

Yes.

> Or do you mean that you've written grammar rules that match syntactically
> incorrect input?

More or less.

The grammar specification has a notion of the 'largest valid subexpression'.
I use the error token to implement that: the error token mark the end of the 
expression.
At this point I call yyerrok and continue. This works fine unless one of the 
largest
valid subexpression is empty because bison will discard token instead fo 
reporting a syntax
error.

Basically I do:
list: expr
    | list error  {if (!once) { yyerrok; } once=1;}  expr {once=0;}
;
the variable once preventing two error to coalesce into one.

> > However, in this contest, discarding
> > tokens is highly undesirable, much more so than aborting.
> 
> Do you mean that you've declared destructors with %destructor?  

No.

However Chris verBurg  suggested to me to add a destructor
%destructor { ++number_discarded_symbols; } 
and to check at the end of parsing if number_discarded_symbols > 1 in which case
the program abort with a syntax error. So far this has been successful.

Cheers,
Bill.