我正在尝试将Boost Spirit X3与语义 Action 一起使用,同时将结构解析为AST。如果我使用的规则没有单独的定义和实例化,那么它可以正常工作,例如:
#include <vector>
#include <string>
#include <iostream>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/home/x3.hpp>
namespace ast
{
struct ast_struct
{
int number;
std::vector<int> numbers;
};
}
BOOST_FUSION_ADAPT_STRUCT(
ast::ast_struct,
(int, number)
(std::vector<int>, numbers)
)
namespace x3 = boost::spirit::x3;
using namespace std;
void parse( const std::string &data )
{
string::const_iterator begin = data.begin();
string::const_iterator end = data.end();
unsigned n(0);
auto f = [&n]( auto &ctx )
{
n = x3::_attr(ctx);
};
ast::ast_struct ast;
bool r = x3::parse( begin, end,
x3::int_[f] >> +( x3::omit[+x3::blank] >> x3::int_ ), ast );
if ( r && begin == end )
{
cout << "n: " << n << ", ";
std::copy(ast.numbers.begin(), ast.numbers.end(),
std::ostream_iterator<int>(std::cout << ast.numbers.size() << " elements: ", " "));
cout << endl;
}
else
cout << "Parse failed" << endl;
}
int main()
{
parse( "3 1 2 3" );
parse( "4 1 2 3 4" );
return 0;
}
运行上面的代码(与标志-std = c++ 14编译)输出预期结果:
n: 3, 3 elements: 1 2 3
n: 4, 4 elements: 1 2 3 4
现在,我试图让我的Spirit X3解析器的组织方式与Boost Boost X3的calc 9 example大致相同,但是它不起作用:
ast.hxx:
#ifndef AST_HXX
#define AST_HXX
#include <vector>
#include <boost/fusion/include/adapt_struct.hpp>
namespace ast
{
struct ast_struct
{
int number;
std::vector<int> numbers;
};
}
BOOST_FUSION_ADAPT_STRUCT(
ast::ast_struct,
(int, number)
(std::vector<int>, numbers)
)
#endif
grammar.hxx:
#ifndef GRAMMAR_HXX
#define GRAMMAR_HXX
#include "ast.hxx"
#include <boost/spirit/home/x3.hpp>
namespace parser
{
namespace x3 = boost::spirit::x3;
using my_rule_type = x3::rule<class my_rule_class, ast::ast_struct>;
BOOST_SPIRIT_DECLARE( my_rule_type );
const my_rule_type &get_my_rule();
}
#endif
grammar.cxx:
#include "grammar_def.hxx"
#include "config.hxx"
namespace parser
{
BOOST_SPIRIT_INSTANTIATE( my_rule_type, iterator_type, context_type )
}
grammar_def.hxx:
#ifndef GRAMMAR_DEF_HXX
#define GRAMMAR_DEF_HXX
#include <iostream>
#include <boost/spirit/home/x3.hpp>
#include "grammar.hxx"
#include "ast.hxx"
namespace parser
{
namespace x3 = boost::spirit::x3;
const my_rule_type my_rule( "my_rule" );
unsigned n;
auto f = []( auto &ctx )
{
n = x3::_attr(ctx);
};
auto my_rule_def = x3::int_[f] >> +( x3::omit[+x3::blank] >> x3::int_ );
BOOST_SPIRIT_DEFINE( my_rule )
const my_rule_type &get_my_rule()
{
return my_rule;
}
}
#endif
config.hxx:
#ifndef CONFIG_HXX
#define CONFIG_HXX
#include <string>
#include <boost/spirit/home/x3.hpp>
namespace parser
{
namespace x3 = boost::spirit::x3;
using iterator_type = std::string::const_iterator;
using context_type = x3::unused_type;
}
#endif
main.cxx:
#include "ast.hxx"
#include "grammar.hxx"
#include "config.hxx"
#include <iostream>
#include <boost/spirit/home/x3.hpp>
#include <string>
namespace x3 = boost::spirit::x3;
using namespace std;
void parse( const std::string &data )
{
parser::iterator_type begin = data.begin();
parser::iterator_type end = data.end();
ast::ast_struct ast;
cout << "Parsing [" << string(begin,end) << "]" << endl;
bool r = x3::parse( begin, end, parser::get_my_rule(), ast );
if ( r && begin == end )
{
std::copy(ast.numbers.begin(), ast.numbers.end(),
std::ostream_iterator<int>(std::cout << ast.numbers.size() << " elements: ", " "));
cout << endl;
}
else
cout << "Parse failed" << endl;
}
int main()
{
parse( "3 1 2 3" );
parse( "4 1 2 3 4" );
return 0;
}
编译main.cxx和grammar.cxx(标志:-std = c++ 14)并运行上面的代码将输出:
Parsing [3 1 2 3]
0 elements:
Parsing [4 1 2 3 4]
0 elements:
我为冗长的源代码表示歉意,并尝试将其尽量缩小。
请注意,我对无符号n全局变量有一些用法,它将与自定义的repeat指令一起使用(请参阅question here和one of the solutions here)。为了使问题集中注意力,我从该问题中删除了重复部分,因此即使在此示例中我可以删除语义 Action ,也不是可行的解决方案。
我将不胜感激,可以帮助您发现此问题,但尚不清楚为什么上面的代码无法正常工作。先感谢您。
最佳答案
我必须承认,实际上重建您的样本对我来说有点麻烦(称我为懒人...)。
但是,我知道答案和使您的生活更简单的窍门。
答案
规则定义上的语义 Action 会禁止自动属性传播。从Qi docs(X3也是如此,但我总是会丢失指向文档的链接):
绝招
您可以使用n
指令注入(inject)状态(在这种情况下为x3::with<>
引用)。这样,您就没有全局 namespace (n
),并且可以使解析器可重入,线程安全等。
这是我的“ list ”,将它们放在一个文件中:
namespace parsing {
x3::rule<struct parser, ast::ast_struct> parser {"parser"};
struct state_tag { };
auto record_number = [](auto &ctx) {
unsigned& n = x3::get<state_tag>(ctx);
n = x3::_attr(ctx);
};
auto parser_def = x3::rule<struct parser_def, ast::ast_struct> {}
%= x3::int_[record_number] >> +(x3::omit[+x3::blank] >> x3::int_);
BOOST_SPIRIT_DEFINE(parser)
}
请注意,
get<state_tag>(ctx)
返回reference_wrapper<unsigned>
只是因为我们使用解析器,如下所示:void parse(const std::string &data) {
using namespace std;
ast::ast_struct ast;
unsigned n;
auto parser = x3::with<parsing::state_tag>(ref(n)) [parsing::parser] >> x3::eoi;
if (x3::parse(data.begin(), data.end(), parser, ast)) {
cout << "n: " << n << ", ";
copy(ast.numbers.begin(), ast.numbers.end(), ostream_iterator<int>(cout << ast.numbers.size() << " elements: ", " "));
cout << "\n";
} else
cout << "Parse failed\n";
}
现场演示Live On Coliru
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/home/x3.hpp>
#include <iostream>
namespace ast {
struct ast_struct {
int number;
std::vector<int> numbers;
};
}
BOOST_FUSION_ADAPT_STRUCT(ast::ast_struct, number, numbers)
namespace x3 = boost::spirit::x3;
namespace parsing {
x3::rule<struct parser, ast::ast_struct> parser {"parser"};
struct state_tag { };
auto record_number = [](auto &ctx) {
unsigned& n = x3::get<state_tag>(ctx); // note: returns reference_wrapper<T>
n = x3::_attr(ctx);
};
auto parser_def = x3::rule<struct parser_def, ast::ast_struct> {}
%= x3::int_[record_number] >> +(x3::omit[+x3::blank] >> x3::int_);
BOOST_SPIRIT_DEFINE(parser)
}
void parse(const std::string &data) {
using namespace std;
ast::ast_struct ast;
unsigned n = 0;
auto parser = x3::with<parsing::state_tag>(ref(n)) [parsing::parser] >> x3::eoi;
if (x3::parse(data.begin(), data.end(), parser, ast)) {
cout << "n: " << n << ", ";
copy(ast.numbers.begin(), ast.numbers.end(), ostream_iterator<int>(cout << ast.numbers.size() << " elements: ", " "));
cout << "\n";
} else
cout << "Parse failed\n";
}
int main() {
parse("3 1 2 3");
parse("4 1 2 3 4");
}
打印n: 3, 3 elements: 1 2 3
n: 4, 4 elements: 1 2 3 4