Accept brackets correctly in bisonc++

Question

I've tried to write a basic syntax checker using bisonc++

The rules are:

expression -> OPEN_BRACKET expression CLOSE_BRACKET
expression -> expression operator expression

operator -> PLUS
operator -> MINUS

If I try to run the compiled code, I get an error at this line:

(a+b)-(c+d)

The first rule is applied, the leftmost and the rightmost brackets are the OPEN_BRACKET and the CLOSE_BRACKET. The remaining expression is: a+b)-(c+d

How is it possible to prevent this behaviour? Is it possible to count the open and closed brackets?

---

Edit

The expression grammar:

expression:
    OPEN_BRACKET expression CLOSE_BRACKET
    {
        //
    }        
| operator
    {
        //
    }        
| VARIABLE
    {
        //
    }

;


operator:
    expression PLUS expression
    {
        //
    }

| expression MINUS expression
    {
        //
    }

;

---

Edit2

The lexer

CHAR  [a-z]
WS    [ \t\n]

%%

{CHAR}+     return Parser::VARIABLE;

"+"         return Parser::PLUS;
"-"         return Parser::MINUS;

"("         return Parser::OPEN_BRACKET;
")"         return Parser::CLOSE_BRACKET;

Answer 1

This is not a normal expression grammar. Try the normal one.

expression
    : term
    | expression '+' term
    | expression '-' term
    ;
term
    : factor
    | term '*' factor
    | term '/' factor
    | term '%' factor
    ;
factor
    : primary
    | '-' factor // unary minus
    | primary '^' factor // exponentiation, right-associative
    ;
primary
    : identifier
    | literal
    | '(' expression ')'
    ;

Note also the above method of indenting and aligning, and that you only have to return yytext[0] from the lexer for single special characters: you don't need special token names, and it's more readable without them:

CHAR [a-zA-Z]
DIGIT [0-9]
WHITESPACE [ \t\r\n]
%%
{CHAR}+       { return Parser::VARIABLE; }
{DIGIT}+      { return Parser::LITERAL; }
{WHITESPACE}+ ;
.             { return yytext[0]; }