4
我希望能夠解析一個數字,以存儲它的原始來源,並跟蹤其來源中的位置,保存它在結構本身。助推精神與源解析
這是我到目前爲止有:
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/spirit/home/support/iterators/line_pos_iterator.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>
#include <iostream>
#include <iomanip>
#include <ios>
#include <string>
#include <complex>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>
struct Position
{
Position()
: line(-1)
{
}
size_t line;
};
struct Number : public Position
{
Number()
: Position()
, value(-1)
, source()
{
}
unsigned value;
std::string source;
};
using namespace boost::spirit;
BOOST_FUSION_ADAPT_STRUCT(Number,
(unsigned, value)
(std::string, source)
(size_t, line)
);
template <typename Iterator>
struct source_hex : qi::grammar<Iterator, Number()>
{
source_hex() : source_hex::base_type(start)
{
using qi::eps;
using qi::hex;
using qi::lit;
using qi::raw;
using qi::_val;
using qi::_1;
using ascii::char_;
namespace phx = boost::phoenix;
using phx::at_c;
using phx::begin;
using phx::end;
using phx::construct;
start = raw[ (lit("0x") | lit("0X"))
>> hex [at_c<0>(_val) = _1]
][at_c<2>(_val) = get_line(begin(_1))]
[at_c<1>(_val) = construct<std::string>(begin(_1), end(_1))]
;
}
qi::rule<Iterator, Number()> start;
};
和測試代碼:
typedef line_pos_iterator<std::string::const_iterator> Iterator;
source_hex<Iterator> g;
Iterator iter(str.begin());
Iterator end(str.end());
Number number;
bool r = parse(iter, end, g, number);
if (r && iter == end) {
std::cout << number.line << ": 0x" << std::setw(8) << std::setfill('0') << std::hex << number.value << " // " << number.source << "\n";
} else
std::cout << "Parsing failed\n";
什麼,我沒有得到就是爲什麼行的迭代器:
[at_c<2>(_val) = get_line(begin(_1))]
不是line_pos_iterator,即使這是我正在使用的解析器。 我會欣賞解釋以及如何解決問題的想法 - 無論如何。
,顯然我在做什麼是完全關閉 - 因爲get_line是語法 – gsf
你需要調用'get_line'爲「懶惰」的建設過程中調用functor(鳳凰男演員)。見[這個答案](http://stackoverflow.com/questions/8358975/cross-platform-way-to-get-line-number-of-an-ini-file-where-given-option-was-foun/ 8365427#8365427)的例子(Inifile解析器),它使用它 – sehe