- {
- yySymbol yytoken_to_shift;
- size_t yys;
- size_t yyn = yystack.yytops.yysize;
-
- /* yyprocessOneStack returns one of three things:
-
- - An error flag. If the caller is yyprocessOneStack, it
- immediately returns as well. When the caller is finally
- yyparse, it jumps to an error label via YYCHK1.
-
- - yyok, but yyprocessOneStack has invoked yymarkStackDeleted
- (&yystack, yys), which sets the top state of yys to NULL. Thus,
- yyparse's following invocation of yyremoveDeletes will remove
- the stack.
-
- - yyok, when ready to shift a token.
-
- Except in the first case, yyparse will invoke yyremoveDeletes and
- then shift the next token onto all remaining stacks. This
- synchronization of the shift (that is, after all preceding
- reductions on all stacks) helps prevents double destructor calls
- on yylval in the event of memory exhaustion. */
-
- for (yys = 0; yys < yyn; yys += 1)
- YYCHK1 (yyprocessOneStack (&yystack, yys, yyposn,
- yylvalp, yyllocp]b4_lpure_args[));
- yyremoveDeletes (&yystack);
- 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, yytoken is set to YYEMPTY
- before the loop to make sure the user destructor for yylval isn't
- called twice. */
- yytoken_to_shift = yytoken;
- yytoken = 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, yylvalp, yyllocp);
- yyglrShift (&yystack, yys, yyaction, yyposn,
- yylvalp, yyllocp);
- YYDPRINTF ((stderr, "Stack %lu now in state #%d\n",
- (unsigned long int) yys,
- yystack.yytops.yystates[yys]->yylrState));
- }
- if (yystack.yytops.yysize == 0)
- {
- yyundeleteLastStack (&yystack);
- if (yystack.yytops.yysize == 0)
- yyFail (&yystack][]b4_lpure_args[, YY_("syntax error"));
- YYCHK1 (yyresolveStack (&yystack]b4_user_args[));
- YYDPRINTF ((stderr, "Returning to deterministic operation.\n"));
-]b4_location_if([[ yystack.yyerror_range[1].yystate.yyloc = *yyllocp;]])[
- yyreportSyntaxError (&yystack, yylvalp, yyllocp]b4_user_args[);
- goto yyuser_error;
- }
- else if (yystack.yytops.yysize == 1)
- {
- YYCHK1 (yyresolveStack (&yystack]b4_user_args[));
- YYDPRINTF ((stderr, "Returning to deterministic operation.\n"));
- yycompressStack (&yystack);
- break;
- }
- }
+ {
+ yySymbol yytoken_to_shift;
+ size_t yys;
+
+ for (yys = 0; yys < yystack.yytops.yysize; yys += 1)
+ yystackp->yytops.yylookaheadNeeds[yys] = yychar != YYEMPTY;
+
+ /* yyprocessOneStack returns one of three things:
+
+ - An error flag. If the caller is yyprocessOneStack, it
+ immediately returns as well. When the caller is finally
+ yyparse, it jumps to an error label via YYCHK1.
+
+ - yyok, but yyprocessOneStack has invoked yymarkStackDeleted
+ (&yystack, yys), which sets the top state of yys to NULL. Thus,
+ yyparse's following invocation of yyremoveDeletes will remove
+ the stack.
+
+ - yyok, when ready to shift a token.
+
+ Except in the first case, yyparse will invoke yyremoveDeletes and
+ then shift the next token onto all remaining stacks. This
+ synchronization of the shift (that is, after all preceding
+ reductions on all stacks) helps prevent double destructor calls
+ on yylval in the event of memory exhaustion. */
+
+ for (yys = 0; yys < yystack.yytops.yysize; yys += 1)
+ YYCHK1 (yyprocessOneStack (&yystack, yys, yyposn]b4_lpure_args[));
+ yyremoveDeletes (&yystack);
+ if (yystack.yytops.yysize == 0)
+ {
+ yyundeleteLastStack (&yystack);
+ if (yystack.yytops.yysize == 0)
+ yyFail (&yystack][]b4_lpure_args[, YY_("syntax error"));
+ YYCHK1 (yyresolveStack (&yystack]b4_user_args[));
+ YYDPRINTF ((stderr, "Returning to deterministic operation.\n"));
+]b4_locations_if([[ yystack.yyerror_range[1].yystate.yyloc = yylloc;]])[
+ yyreportSyntaxError (&yystack]b4_user_args[);
+ goto yyuser_error;
+ }
+
+ /* 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 < yystack.yytops.yysize; 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]b4_locations_if([, &yylloc])[);
+ YYDPRINTF ((stderr, "Stack %lu now in state #%d\n",
+ (unsigned long int) yys,
+ yystack.yytops.yystates[yys]->yylrState));
+ }
+
+ if (yystack.yytops.yysize == 1)
+ {
+ YYCHK1 (yyresolveStack (&yystack]b4_user_args[));
+ YYDPRINTF ((stderr, "Returning to deterministic operation.\n"));
+ yycompressStack (&yystack);
+ break;
+ }
+ }
projects / bison.git / blobdiff