Re[2]: LALR : How it work during parsing time with empty rule ? what to do to the stack?

[Baldurien (club internet)]

Bonjour Hans,

Le mardi 13 septembre 2005 à 00:45:16, vous écriviez :

HA> On 12 Sep 2005, at 23:31, Baldurien (club internet) wrote:

>> I've  made  a  little  parser in PHP based on bison output file, and I
>> experiment a little problem :
>>
>> When I have to reduce, saying by rule X : Y '+' X I know I have to pop
>> 3  times, but what happens when the rule is simply X : /* empty */ ? I
>> can't pop at all, so what should I do?

HA> Just don't pop. If  s  is the state on top of the state stack, the
HA> total effect is that the state goto(s, X) is pushed onto it. If you
HA> have a stack for non-terminals and terminals, which the Bison  
HA> generated parser doesn't, then X will be pushed onto that. The rule
HA> action is executed.

HA>    Hans Aberg


So, in term of coding I've got these (in php) :

$reduce: the rule we use to reduce.
$ss : the stack of state
$sl : the level of this stack. The top state is $ss[$sl-1].
$ls  :  the  location stack, it's a stack that contains only the start
and  the  end  line of a rule. I use it in order to have comprehensive
error  message (like : syntax error between line x and y for a string,
etc).

$rule = $__xtf_parser_rule_lhs[$reduce];

/**
 * We have to pop a certain amount of time
 */
$pop = $__xtf_parser_rule_count[$reduce];
$sl -= $pop;

$goto = &$__xtf_parser_goto[$__xtf_parser_sgoto[$ss[$sl-1]]];

/* call reduce handler, etc : action set by the user */
        
$ls[$sl-1] = array(
  $ls[$sl-1][0],
  $ls[$sl-2+$pop][1]
);

$ss[$sl++] = $goto[$rule];

Here I do $sl++, meaning I increment the stack level (but after). When
pop is null, should it be only $ss[$sl-1] ?


-- 
Cordialement,
 Baldurien                            Courriel : address@hidden