C++ 从子规则中指定值
我试图解析2种不同类型的字符串,并将值赋给结构。对于性能,我尝试使用boost spirit子规则 字符串可以是以下类型C++ 从子规则中指定值,c++,parsing,boost,boost-spirit,boost-spirit-qi,C++,Parsing,Boost,Boost Spirit,Boost Spirit Qi,我试图解析2种不同类型的字符串,并将值赋给结构。对于性能,我尝试使用boost spirit子规则 字符串可以是以下类型 Animal Type | Animal Attributes Ex DOG | Name=tim | Barks=Yes | Has a Tail=N | Address=3 infinite loop BIRD| Name=poc | Tweets=Yes| Address=10 stack overflow street 这些值存储在下面的Dog
Animal Type | Animal Attributes
Ex
DOG | Name=tim | Barks=Yes | Has a Tail=N | Address=3 infinite loop
BIRD| Name=poc | Tweets=Yes| Address=10 stack overflow street
这些值存储在下面的Dog和Bird结构数组中
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/repository/include/qi_subrule.hpp>
#include <boost/spirit/include/qi_symbols.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>
#include <iostream>
using std::cout;
using std::endl;
using std::cerr;
struct Dog
{
std::string Name;
bool Barks;
bool HasATail;
std::string Address;
};
struct Bird
{
std::string Name;
bool Tweets;
std::string Address;
};
namespace qi = boost::spirit::qi;
namespace repo = boost::spirit::repository;
namespace ascii = boost::spirit::ascii;
namespace phx = boost::phoenix;
template <typename Iterator>
struct ZooGrammar : public qi::grammar<Iterator, ascii::space_type>
{
ZooGrammar() : ZooGrammar::base_type(start_)
{
using qi::char_;
using qi::lit_;
using qi::_1;
using boost::phoenix::ref;
boost::spirit::qi::symbols<char, bool> yesno_;
yesno_.add("Y", true)("N", false);
start_ = (
dog_ | bird_,
dog_ = "DOG" >> lit_[ref(d.Name) = _1]>> '|'
>>"Barks=">>yesno_[ref(d.Barks) = _1] >>'|'
>>"Has a Tail=">>yesno_[ref(d.HasATail) = _1] >> '|'
>>lit_[ref(d.Address) = _1]
,
bird_ = "BIRD" >> lit_[ref(b.Name) = _1]>> '|'
>>"Tweets=">>yesno_[ref(b.Tweets) = _1] >>'|'
>>lit_[ref(b.Address) = _1]
);
}
qi::rule<Iterator, ascii::space_type> start_;
repo::qi::subrule<0> dog_;
repo::qi::subrule<1> bird_;
Bird b;
Dog d;
};
int main()
{
std::string test1="DOG | Name=tim | Barks=Yes | Has a Tail=N | Address=3 infinite loop";
std::string test2="BIRD| Name=poc | Tweets=Yes| Address=10 stack overflow street";
using boost::spirit::ascii::space;
typedef std::string::const_iterator iterator_type;
typedef ZooGrammar<iterator_type> grammar;
iterator_type start = test1.begin();
iterator_type end = test1.end();
ZooGrammar g;
if(boost::spirit::qi::phrase_parse(start, end, g, space))
{
cout<<"matched"<<endl;
}
}
#定义提升(精神)使用(凤凰)
#包括
#包括
#包括
#包括
#包括
#包括
#包括
使用std::cout;
使用std::endl;
使用std::cerr;
结构狗
{
std::字符串名;
狗叫;
布尔·哈萨泰尔;
std::字符串地址;
};
结构鸟
{
std::字符串名;
布尔推特;
std::字符串地址;
};
名称空间qi=boost::spirit::qi;
名称空间repo=boost::spirit::repository;
名称空间ascii=boost::spirit::ascii;
名称空间phx=boost::phoenix;
模板
结构zoogram:public qi::grammar
{
ZooGrammar():ZooGrammar::基本类型(开始类型)
{
使用qi::char\ux;
使用qi::lit_2;;
使用气::_1;
使用boost::phoenix::ref;
精神:气:符号是的;
是的,没有。加上(“Y”,正确)(“N”,错误);
开始时间=(
狗|鸟|,
dog=“dog”>>lit.[ref(d.Name)=\u1]>>“|”
>>“吠声=“>>是”否[ref(d.Barks)=\u 1]>>“|”
>>“有一条尾巴=“>>是的”[ref(d.HasATail)=\u 1]>>“|”
>>lit_uu[ref(d.Address)=1]
,
bird=“bird”>>lit.[ref(b.Name)=\u 1]>>“|”
>>“Tweets=“>>yesno”[ref(b.Tweets)=\u 1]>>“|”
>>lit_uu[ref(b.地址)=1]
);
}
qi::规则开始;
回购协议::qi::子规则;
回购协议::合格中介机构::子规则;
鸟b;
狗d;
};
int main()
{
std::string test1=“DOG | Name=tim | Barks=Yes | Has a Tail=N | Address=3无限循环”;
std::string test2=“BIRD | Name=poc | Tweets=Yes | Address=10堆栈溢出街”;
使用boost::spirit::ascii::space;
typedef std::string::const_iterator iterator_type;
类型语法;
迭代器_type start=test1.begin();
迭代器类型end=test1.end();
动物语法;
if(boost::spirit::qi::短语解析(开始、结束、g、空格))
{
子规则有点过时了。老实说,我甚至不知道精神V2里还有这样的东西
我建议使用常规的Spirit V2属性传播,这会让事情一下子变得更具可读性:
dog_ = qi::lit("DOG") >> '|' >> "Name=" >> lit_ >> '|'
>> "Barks=" >> yesno_ >> '|'
>> "Has a Tail=" >> yesno_ >> '|'
>> "Address=" >> lit_
;
bird_ = qi::lit("BIRD") >> '|' >> "Name=" >> lit_ >> '|'
>> "Tweets=" >> yesno_ >> '|'
>> "Address=" >> lit_
;
start_ = dog_ | bird_;
我想象了一个lit\u
规则(因为qi::lit\u
没有敲响任何钟声):
当然,您需要调整属性类型,只要它们没有内置的支持(与boost::variant
、std::string
和bool
一样,它们都是在不使用任何附加代码的情况下处理的):
现在扩展程序以打印一些调试信息,输出为:
Matched: [DOG|Name=tim |Barks=Yes|Has a Tail=No|Address=3 infinite loop]
Matched: [BIRD|Name=poc |Tweets=Yes|Address=10 stack overflow street]
完整示例代码
/#定义BOOST_SPIRIT_调试
#定义增强\u精神\u使用\u凤凰\u V3
#包括
#包括
#包括
静态常量char*YesNo(bool b){返回b?“是”:“否”}
结构狗{
std::字符串名;
狗叫;
布尔·哈萨泰尔;
std::字符串地址;
friend std::ostream&运算符“|”
>>“Address=“>>点亮_
;
bird_uuqi::lit(“bird”)>>'|'>>“Name=“>>lit_u>>'|”
>>“Tweets=“>>是”否“|”
>>“Address=“>>点亮_
;
开始=狗|鸟|;
lit_uqi=lexeme[*~qi::char_qi('|')];
BOOST_SPIRIT_DEBUG_节点((dog_)(bird_)(start_)(lit_))
}
私人:
qi::规则开始;
齐:规则点燃;
齐:统治狗;
齐:规则鸟;
qi::符号是的;
};
int main()
{
typedef std::string::const_iterator iterator_type;
类型语法;
对于(std::string常量输入:{
“狗|名字=蒂姆|吠叫=是|有一条尾巴=N |地址=3无限循环”,
“BIRD | Name=poc | Tweets=Yes | Address=10 stack overflow street”
})
{
迭代器_type start=input.begin();
迭代器_type end=input.end();
语法g;
动物;
if(qi::短语解析(开始、结束、g、ascii::空格、动物))
std::cout。子规则似乎已被放弃,无法使用现代编译器使用c++11进行编译,即使它使用c++98/03进行编译,也无法工作。我不知道您是否意识到,我认为这将有助于您的解析器。
BOOST_FUSION_ADAPT_STRUCT(Dog,
(std::string, Name)(bool, Barks)(bool, HasATail)(std::string, Address))
BOOST_FUSION_ADAPT_STRUCT(Bird,
(std::string, Name)(bool, Tweets)(std::string, Address))
Matched: [DOG|Name=tim |Barks=Yes|Has a Tail=No|Address=3 infinite loop]
Matched: [BIRD|Name=poc |Tweets=Yes|Address=10 stack overflow street]
//#define BOOST_SPIRIT_DEBUG
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/qi_symbols.hpp>
static const char* YesNo(bool b) { return b?"Yes":"No"; }
struct Dog {
std::string Name;
bool Barks;
bool HasATail;
std::string Address;
friend std::ostream& operator <<(std::ostream& os, Dog const& o) {
return os << "[DOG|Name=" << o.Name << "|Barks=" << YesNo(o.Barks) << "|Has a Tail=" << YesNo(o.HasATail) << "|Address=" << o.Address << "]";
}
};
struct Bird {
std::string Name;
bool Tweets;
std::string Address;
friend std::ostream& operator <<(std::ostream& os, Bird const& o) {
return os << "[BIRD|Name=" << o.Name << "|Tweets=" << YesNo(o.Tweets) << "|Address=" << o.Address << "]";
}
};
typedef boost::variant<Dog, Bird> ZooAnimal;
BOOST_FUSION_ADAPT_STRUCT(Dog, (std::string, Name)(bool, Barks)(bool, HasATail)(std::string, Address))
BOOST_FUSION_ADAPT_STRUCT(Bird, (std::string, Name)(bool, Tweets)(std::string, Address))
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
template <typename Iterator>
struct ZooGrammar : public qi::grammar<Iterator, ZooAnimal(), ascii::space_type>
{
ZooGrammar() : ZooGrammar::base_type(start_)
{
using qi::_1;
yesno_.add("Yes", true)("Y", true)("No", false)("N", false);
dog_ = qi::lit("DOG") >> '|' >> "Name=" >> lit_ >> '|'
>> "Barks=" >> yesno_ >> '|'
>> "Has a Tail=" >> yesno_ >> '|'
>> "Address=" >> lit_
;
bird_ = qi::lit("BIRD") >> '|' >> "Name=" >> lit_ >> '|'
>> "Tweets=" >> yesno_ >> '|'
>> "Address=" >> lit_
;
start_ = dog_ | bird_;
lit_ = qi::lexeme [ *~qi::char_('|') ];
BOOST_SPIRIT_DEBUG_NODES((dog_)(bird_)(start_)(lit_))
}
private:
qi::rule<Iterator, ZooAnimal(), ascii::space_type> start_;
qi::rule<Iterator, std::string(), ascii::space_type> lit_;
qi::rule<Iterator, Dog(), ascii::space_type> dog_;
qi::rule<Iterator, Bird(), ascii::space_type> bird_;
qi::symbols<char, bool> yesno_;
};
int main()
{
typedef std::string::const_iterator iterator_type;
typedef ZooGrammar<iterator_type> grammar;
for (std::string const input : {
"DOG | Name=tim | Barks=Yes | Has a Tail=N | Address=3 infinite loop",
"BIRD| Name=poc | Tweets=Yes| Address=10 stack overflow street"
})
{
iterator_type start = input.begin();
iterator_type end = input.end();
grammar g;
ZooAnimal animal;
if(qi::phrase_parse(start, end, g, ascii::space, animal))
std::cout << "Matched: " << animal << "\n";
else
std::cout << "Parse failed\n";
if (start != end)
std::cout << "Remaining input: '" << std::string(start, end) << "'\n";
}
}