如何使用 Boost Spirit 函式庫解析 C 語言中的布林運算式？

#include <boost/spirit/include/qi.hpp>
#include <boost/variant/recursive_wrapper.hpp>

using namespace boost::spirit;
using namespace boost::spirit::qi;

struct op_or  {};
struct op_and {};
struct op_xor {};

typedef std::string var;
template <typename tag> struct binop;
template <typename tag> struct unop;

typedef boost::variant<var,
        boost::recursive_wrapper<unop <op_not> >, 
        boost::recursive_wrapper<binop<op_and> >,
        boost::recursive_wrapper<binop<op_xor> >,
        boost::recursive_wrapper<binop<op_or> >
        > expr;

template <typename tag> struct binop 
{ 
    explicit binop(const expr&amp; l, const expr&amp; r) : oper1(l), oper2(r) { }
    expr oper1, oper2; 
};

template <typename tag> struct unop  
{ 
    explicit unop(const expr&amp; o) : oper1(o) { }
    expr oper1; 
};

template <typename It, typename Skipper = qi::space_type>
    struct parser : qi::grammar<It, expr(), Skipper>
{
    parser() : parser::base_type(expr_)
    {
        using namespace qi;

        expr_  = or_.alias();
        not_ = ("not" > simple       ) [ _val = phx::construct<unop <op_not> >(_1)     ] | simple [ _val = _1 ];
        or_  = (xor_ >> "or"  >> or_ ) [ _val = phx::construct<binop<op_or> >(_1, _2) ] | xor_   [ _val = _1 ];
        xor_ = (and_ >> "xor" >> xor_) [ _val = phx::construct<binop<op_xor> >(_1, _2) ] | and_   [ _val = _1 ];
        and_ = (not_ >> "and" >> and_) [ _val = phx::construct<binop<op_and> >(_1, _2) ] | not_   [ _val = _1 ];

        simple = (('(' > expr_ > ')') | var_);
        var_ = qi::lexeme[ +alpha ];
    }
  private:
    qi::rule<It, var() , Skipper> var_;
    qi::rule<It, expr(), Skipper> not_, and_, xor_, or_, simple, expr_;
};

auto expr = "a and b xor (c and d or a and b);";
expr parsed_expr;
qi::parse(expr.begin(), expr.end(), parser<std::string::const_iterator>(), parsed_expr);
std::cout << "Parsed expression: " << parsed_expr << std::endl;

Parsed expression: ((a and b) xor ((c and d) or (a and b)))