why does $1 in yacc/bison has a value of 0 -

i have following production in bison spec:

op : '+' { printf("%d %d %c\n", $1, '+', '+'); } 

when input + following output:

0 43 + 

can explain why $1 has value of 0, shouldn't 43? missing?

edit

there no flex file, can provide bison grammar:

%{ #include <stdio.h> #include <ctype.h> #include <string.h>  int yylex(); int yyerror();  %}  %token number  %%  lexp : number       | '(' op lexp-seq ')'      ; op : '+' { printf("%d %d %c\n", $1, '+', '+'); }    | '-' { printf("%d %d %c\n", $1, '-', '-'); }    | '*' { printf("%d %d %c\n", $1, '*', '*'); }    ;  lexp-seq : lexp-seq lexp          | lexp          ;  %%  int main(int argc, char** argv) {   if (2 == argc && (0 == strcmp("-g", argv[1])))     yydebug = 1;    return yyparse(); }  int yylex() {   int c;    /* eliminate blanks*/   while((c = getchar()) == ' ');    if (isdigit(c)) {     ungetc(c, stdin);     scanf("%d", &yylval);     return (number);   }    /* makes parse stop */   if (c == '\n') return 0;    return (c); }  int yyerror(char * s) {   fprintf(stderr, "%s\n", s);   return 0; } /* allows printing of error message */ 

$1 semantic value of first symbol on right-hand side, in case '+'. since terminal, semantic value whatever value of yylval when scanner returned '+' token parser.

since scanner not set yylval in case returns '+' (which totally normal), use of $1 in production not defined. normally, grammar doesn't reference semantic values of tokens '+', purely syntactic , don't have semantic values.

however, since yylval static variable, have been initialized 0, continue have value until set (for example, while scanning number).