解析表达式时Boost :: Spirit段错误

Question

尝试使用Boost :: Spirit解析表达式时遇到段错误。

我将代码简化为只是一个操作的int表达式类型。示例中的括号使代码段错误。

我最初处理了括号，但我逐步删除了代码以获得最简单的示例。是不是语法应该处理未知的括号而只是解析失败？

#include <string>
#include <iostream>
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/home/qi.hpp>
#include <boost/variant/apply_visitor.hpp>
#include <boost/variant/recursive_variant.hpp>
#include <boost/fusion/include/adapt_struct.hpp>

namespace ascii = boost::spirit::ascii;
namespace qi = boost::spirit::qi;
using std::string;

namespace ast {

struct operation;

using expression = boost::variant<int, boost::recursive_wrapper<operation>>;

struct operation {
  expression lhs_;
  string operator_;
  expression rhs_;
};
}

BOOST_FUSION_ADAPT_STRUCT(ast::operation, (ast::expression, lhs_),
                          (string, operator_), (ast::expression, rhs_))

namespace parser {
template <typename Iterator>
struct language_grammar
    : qi::grammar<Iterator, ast::expression(), ascii::space_type> {
  language_grammar() : language_grammar::base_type(expression) {
    expression = qi::int_ | operation;
    operation = expression >> operator_ >> expression;
    operator_ = "+";
  }
  qi::rule<Iterator, ast::expression(), ascii::space_type> expression;
  qi::rule<Iterator, string()> operator_;
  qi::rule<Iterator, ast::operation(), ascii::space_type> operation;
};
}

int main() {
  string to_parse = "(1 + 2)";
  ast::expression expr;
  parser::language_grammar<string::const_iterator> grammar;
  string::const_iterator iter = to_parse.begin();
  string::const_iterator end = to_parse.end();
  ascii::space_type space;
  bool r = phrase_parse(iter, end, grammar, space, expr);
  std::cout << r << std::endl;
  return 0;
}