+ yyn = yystack.yytops.yysize;
+
+ /* If any yyglrShift call fails, it will fail after shifting. Thus,
+ a copy of yylval will already be on stack 0 in the event of a
+ failure in the following loop. Thus, yychar is set to YYEMPTY
+ before the loop to make sure the user destructor for yylval isn't
+ called twice. */
+ yytoken_to_shift = YYTRANSLATE (yychar);
+ yychar = YYEMPTY;
+ yyposn += 1;
+ for (yys = 0; yys < yyn; yys += 1)
+ {
+ int yyaction;
+ const short int* yyconflicts;
+ yyStateNum yystate = yystack.yytops.yystates[yys]->yylrState;
+ yygetLRActions (yystate, yytoken_to_shift, &yyaction,
+ &yyconflicts);
+ /* Note that yyconflicts were handled by yyprocessOneStack. */
+ YYDPRINTF ((stderr, "On stack %lu, ", (unsigned long int) yys));
+ YY_SYMBOL_PRINT ("shifting", yytoken_to_shift, &yylval, &yylloc);
+ yyglrShift (&yystack, yys, yyaction, yyposn,
+ &yylval, &yylloc);
+ YYDPRINTF ((stderr, "Stack %lu now in state #%d\n",
+ (unsigned long int) yys,
+ yystack.yytops.yystates[yys]->yylrState));
+ }