Reputation: 4528
I have a following production in a bison
spec:
op : '+' { printf("%d %d %c\n", $1, '+', '+'); }
When I input a +
I get the following output:
0 43 +
Can someone explain why $1
has a value of 0, shouldn't it be 43? What am I missing?
EDIT
There is no flex file, but I can provide a 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 the parse stop */
if (c == '\n') return 0;
return (c);
}
int yyerror(char * s) {
fprintf(stderr, "%s\n", s);
return 0;
} /* allows for printing of an error message */
Upvotes: 1
Views: 2740
Reputation: 241861
$1
is the semantic value of the first symbol on the right-hand side, which in this case is '+'
. Since that is a terminal, its semantic value will be whatever the value of yylval
was when the scanner returned a '+'
token to the parser.
Since your scanner does not set yylval
in the case that it returns '+'
(which is totally normal), the use of $1
in that production is not well defined. Normally, a grammar doesn't reference the semantic values of tokens like '+'
, which are purely syntactic and don't have semantic values.
However, since yylval
is a static variable, it will have been initialized to 0, so it will continue to have that value until it is set (for example, while scanning a NUMBER
).
Upvotes: 4