Bison Stack emantic value - debugging

Question

This one is Lexical Analyzer using Flex.
#include <iostream>
#include <cstdio>
#define YY_DECL extern "C" int yylex()
#include "conv.tab.h"
using namespace std;
%}
eq [ \t]*=
%%
[ \t] ;
(?:POINT|LINE) { yylval.ename = strdup(yytext); return ENAME; }
x{eq} { yylval.xval = atof(yytext);
return XVAL; }
y{eq} { yylval.yval = atof(yytext);
return YVAL; }
. ;
%%
And other file is Bison grammar file
%{
#include <iostream>
#include <cstdio>
#include <stdio.h>
using namespace std;
extern "C" int yylex ();
extern "C" int yyparse (void);
extern "C" FILE *yyin;
extern int line_no;
void yyerror(const char *s);
%}
%union{
float xval;
float yval;
char *ename;
}
%token <ename> ENAME
%token XVAL
%token YVAL
%%
converter:
converter ENAME { cout << "entity = " << $2 << endl; }
| converter XVAL {// x -> xval = $2;
cout << "x value = " << endl; }
| converter YVAL {// y -> yval = $2;
cout << "y value = " << endl; }
| ENAME { cout << "entity = " << $1 << endl; }
| XVAL { cout << "xvalue " << endl; }
| YVAL { cout << "yvalue " << endl; }
%%
main() {
FILE *myfile = fopen("conv.aj", "r");
if (!myfile) {
cout << "I can't open file" << endl;
return -1;
}
yyin = myfile;
do{
yydebug = 1;
yyparse();
} while (!feof(yyin));
yydebug = 2;
}
void yyerror(const char *s) {
cout << "Parser error! Message: " << s << endl;
exit(-1);
}
Actually, I want to retrieve values from a file. I used the Bison Debugger and get to know that those values are not able to push onto Bison Stack. So basically I want to push those values onto the stack.My file is like :
POINT
x=38
y=47

Answer 1

Nothing in your lexical analyzer matches a number, so the 38 and 47 from the input will both be handled by your default rule (. ;) which will cause them to be ignored. In your rules for XVAL and YVAL, you call atoi on yytext, which will be x= (or y=); that is clearly not a number and atoi will probably return 0.
It's not clear to me what you mean by "those values are not able to push onto Bison Stack", but I think this problem has nothing to do with bison or its stack.
By the way:
There is no need to have two different members in your semantic type for xval and yval. The type is a union, not a struct, so having two members of the same type (float) is redundant.
flex doesn't do regex captures. So there is really no point avoiding a capture with (?:...); it just obscures your grammar. You might as well use:
POINT|LINE: { yylval.ename = strdup(yytext); return ENAME; }
On the other hand, you might be better off defining two different token types, which would avoid the need for the strdup. (You don't seem to be freeing the duplicated string, so the strdup is also a memory leak.) Alternatively, you could use an enumerated value in your semantic type:
POINT { yylval.ename_enum=POINT; return ENAME; }
LINE { yylval.ename_enum=LINE; return ENAME; }
. ; is not really a good idea, especially during development, because it hides errors (such as the one you have). You can use %option nodefault to avoid flex's default rule, and then flex will present an error when an illegal character is detected.
Unless you're using really old versions of bison and flex, you can just compile the generated code as c++. There should not be a need to use extern "C"