标签:boost-spirit-qi c boost boost-spirit
我想解析特殊结构并将其余部分抛弃.但我不想使用船长.
我想获得这些结构的向量,所以我使用Kleene Star解析器作为主要规则.但是,每次丢弃某些东西时,都会在向量中插入一个默认的构造元素.
这是一个组成的例子.它只是寻找字符串Test并将其余部分抛弃,至少这是计划.但每次规则垃圾成功时,它会在规则all中向向量添加一个默认构造项,输出为7 insteat of 1.如果规则项成功,我怎么能告诉Spirit只添加到向量?
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iostream>
#include <string>
#include <vector>
namespace qi = boost::spirit::qi;
struct container {
std::string name;
bool dummy;
};
BOOST_FUSION_ADAPT_STRUCT(::container,
(std::string, name)
(bool, dummy))
int main() {
typedef std::string::const_iterator iterator;
qi::rule<iterator, std::vector<container>()> all;
qi::rule<iterator, container()> item;
qi::rule<iterator, std::string()> string_rule;
qi::rule<iterator> garbage;
all = *(garbage | item);
garbage = qi::char_ - qi::lit("Test");
string_rule = qi::string("Test");
item = string_rule >> qi::attr(true);
std::vector<container> ast;
std::string input = "blaTestbla";
iterator first = input.begin();
iterator last = input.end();
bool result = qi::parse(first, last, all, ast);
if (result) {
result = first == last;
}
if (result) {
std::cout << "Parsed " << ast.size() << " element(s)" << std::endl;
} else {
std::cout << "failure" << std::endl;
}
}
解决方法:
由于sehe的答案或多或少用于教育目的,我们现在有几个解决方案:
*garbage >> -(item % *garbage) >> *garbage
*garbage >> *(item >> *garbage)
all = *(garbage | item[phx::push_back(qi::_val,qi::_1)]);
来自cv_and_he的解决方案:
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <iostream>
#include <string>
#include <vector>
namespace qi = boost::spirit::qi;
struct container {
std::string name;
bool dummy;
};
BOOST_FUSION_ADAPT_STRUCT(::container,
(std::string, name)
(bool, dummy))
struct container_vector { //ADDED
std::vector<container> data;
};
namespace boost{ namespace spirit{ namespace traits //ADDED
{
template <>
struct is_container<container_vector> : boost::mpl::true_ {};
template <>
struct container_value<container_vector> {
typedef optional<container> type;
};
template <>
struct push_back_container<container_vector,optional<container> > {
static bool call(container_vector& cont, const optional<container>& val) {
if(val)
cont.data.push_back(*val);
return true;
}
};
}}}
int main() {
typedef std::string::const_iterator iterator;
qi::rule<iterator, container_vector()> all; //CHANGED
qi::rule<iterator, container()> item;
qi::rule<iterator, std::string()> string_rule;
qi::rule<iterator> garbage;
all = *(garbage | item);
garbage = qi::char_ - qi::lit("Test");
string_rule = qi::string("Test");
item = string_rule >> qi::attr(true);
container_vector ast; //CHANGED
std::string input = "blaTestbla";
iterator first = input.begin();
iterator last = input.end();
bool result = qi::parse(first, last, all, ast);
if (result) {
result = first == last;
}
if (result) {
std::cout << "Parsed " << ast.data.size() << " element(s)" << std::endl; //CHANGED
} else {
std::cout << "failure" << std::endl;
}
}
虽然我不想使用船长,但我最终得到了:
start = qi::skip(garbage.alias())[*item];
在使用我的生产规则的Linux内核的c文件的不科学测试中,最后一个解决方案是最快的(1-2%).
标签:boost-spirit-qi,c,boost,boost-spirit 来源: https://codeday.me/bug/20190831/1774899.html
本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享; 2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关; 3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关; 4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除; 5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。