我正在尝试为填充抽象语法树的算术表达式编写精神解析器。如果我不尝试填充 AST,解析器会编译,但在当前版本中会失败(有一个 24K 错误)。我正在使用带有 -std=c++11 的 clang++ 3.5.0,并在 Ubuntu 14.4 上运行。
#include <string>
#include <vector>
#include <utility>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>
#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
using std::string;
using std::vector;
using std::pair;
using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;
struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef pair<char, RWTerm> OpAndRWTerm;
typedef pair<RWTerm, vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef pair<char, Factor> OpAndFactor;
struct Term : public pair<Factor, vector<OpAndFactor> >{};
template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
formula_parser() : formula_parser::base_type(expr_rule) {
using boost::spirit::qi::double_;
using boost::spirit::ascii::char_;
factor_rule %= double_ | parenthesis_rule;
parenthesis_rule %= '(' >> expr_rule >> ')';
op_and_factor_rule %= char_("/*") >> factor_rule;
term_rule %= factor_rule >> *op_and_factor_rule;
op_and_term_rule %= char_("+-") >> term_rule;
expr_rule %= term_rule >> *op_and_term_rule;
}
rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
rule<It, Expr(), space_type> expr_rule;
rule<It, OpAndFactor(), space_type> op_and_factor_rule;
rule<It, RWTerm(), space_type> term_rule;
rule<It, Expr(), space_type> parenthesis_rule;
rule<It, Factor(), space_type> factor_rule;
};
int main() {
formula_parser<string::const_iterator> grammar;
}
我从错误消息中了解到,融合在规则 term_rule 中混淆了 Types Factor 和 RWTerm。
我做错了什么?
最佳答案
如果我改变两件事,它会为我编译:
自
Term
继承自std::pair
,Term
是一种新类型。为此,您需要申请BOOST_FUSION_ADAPT_STRUCT
至Term
, 无论这是否已经为std::pair
完成在<boost/fusion/adapted/std_pair.hpp>
:BOOST_FUSION_ADAPT_STRUCT( Term, (Factor, first) (std::vector<OpAndFactor>, second) )
或者,您可以制作
Term
具有两个成员的独立结构,然后应用BOOST_FUSION_ADAPT_STRUCT
对此:struct Term { Factor f; std::vector<OpAndFactor> o;}; BOOST_FUSION_ADAPT_STRUCT( Term, (Factor, f) (std::vector<OpAndFactor>, o) )
顺便说一句:你必须完全符合条件
std::vector
在这里,因为以下 will not compile :using std::vector; BOOST_FUSION_ADAPT_STRUCT( Term, (Factor, f) (vector<OpAndFactor>, o) )
使用
Term
而不是RWTerm
声明term_rule
时:rule<It, Term(), space_type> term_rule;
完整代码:
#include <string>
#include <vector>
#include <utility>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/adapted.hpp>
#include <boost/fusion/include/adapted.hpp>
#include <boost/variant/variant.hpp>
#include <boost/variant/recursive_wrapper.hpp>
using boost::spirit::qi::grammar;
using boost::spirit::qi::space_type;
using boost::spirit::qi::rule;
struct Term; // forward dec
typedef boost::recursive_wrapper<Term> RWTerm;
typedef std::pair<char, RWTerm> OpAndRWTerm;
typedef std::pair<RWTerm, std::vector<OpAndRWTerm> > Expr;
typedef boost::variant<Expr, double> Factor;
typedef std::pair<char, Factor> OpAndFactor;
struct Term : public std::pair<Factor, std::vector<OpAndFactor> >{};
BOOST_FUSION_ADAPT_STRUCT(
Term,
(Factor, first)
(std::vector<OpAndFactor>, second)
)
template<typename It>
struct formula_parser : grammar<It, Expr(), space_type> {
formula_parser() : formula_parser::base_type(expr_rule) {
using boost::spirit::qi::double_;
using boost::spirit::ascii::char_;
factor_rule %= double_ | parenthesis_rule;
parenthesis_rule %= '(' >> expr_rule >> ')';
op_and_factor_rule %= char_("/*") >> factor_rule;
term_rule %= factor_rule >> *op_and_factor_rule;
op_and_term_rule %= char_("+-") >> term_rule;
expr_rule %= term_rule >> *op_and_term_rule;
}
rule<It, OpAndRWTerm(), space_type> op_and_term_rule;
rule<It, Expr(), space_type> expr_rule;
rule<It, OpAndFactor(), space_type> op_and_factor_rule;
rule<It, Term(), space_type> term_rule;
rule<It, Expr(), space_type> parenthesis_rule;
rule<It, Factor(), space_type> factor_rule;
};
int main() {
formula_parser<std::string::const_iterator> grammar;
}
关于c++ - boost::spirit 算术公式解析器编译失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33433911/