将规则生产绑定到我的结构成员时出现编译错误

Getting compilation errors while binding a rule production to my struct members

本文关键字：成员 错误 编译 结构 我的 规则 绑定      更新时间：2023-10-16

使用 Phoenix 绑定编写 Qi 语法时，我遇到了一个编译错误，例如

boost/spirit/home/support/context.hpp(180)：错误 C2338：index_is_out_of_bounds

这里

>> ruleHandId_[phx::bind(&parseContext::handId_, qi::_r1) = qi::_1];

我只是没有太多的凤凰绑定到期，但 perv 绑定在线

ruleStart_ = ruleEncoding_[phx::bind(&parseContext::encoding_, qi::_r1) = qi::_1]

工作正常，没有编译错误

这一切都在VS2013的MSVC下，提升1.56 x86

我在编译错误的代码下做错了什么？

源代码

#include <boost/spirit/include/qi.hpp>
#include <boost/phoenix/phoenix.hpp>
#include <boost/shared_ptr.hpp>
#include <sstream>
namespace sp = boost::spirit;
namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;
struct parseContext {
std::string encoding_;
uint64_t    handId_;
};
typedef boost::shared_ptr<parseContext> parseContextShPtr;
template <typename Iterator>
struct parseGrammar : qi::grammar<Iterator, void(parseContext&)> {
parseGrammar() : parseGrammar::base_type(ruleStart_)
{
ruleStart_ = ruleEncoding_[phx::bind(&parseContext::encoding_, qi::_r1) = qi::_1]
>> ruleHandHeader_;
ruleEncoding_ = qi::lit("ABC");
ruleHandHeader_ = qi::lit("DEF") >> qi::space
>> qi::lit("XYZ #")
>> ruleHandId_[phx::bind(&parseContext::handId_, qi::_r1) = qi::_1];
ruleHandId_ = qi::long_long;
}
// Rules
qi::rule<Iterator, void(parseContext&)> ruleStart_;
qi::rule<Iterator, std::string()> ruleEncoding_;
qi::rule<Iterator> ruleHandHeader_;
qi::rule<Iterator, uint64_t> ruleHandId_;
};
void test()
{
std::string s("ABCDEF XYZ #555: PQI #777");
std::stringstream sb;
sb.unsetf(std::ios::skipws);
sb << s;
const parseGrammar<sp::istream_iterator> p;
sp::istream_iterator b(sb);
sp::istream_iterator e;
parseContextShPtr ctx(new parseContext);
bool r = qi::parse(b, e, p(phx::ref(*ctx.get())));
if (r) {
std::cout << "Success" << std::endl;
}
else {
std::cout << "Failure" << std::endl;
}
std::cout << std::string(b, e).substr(0, 32) << std::endl;
}

某些占位符无法绑定。

这可能是因为ruleEncoding_没有公开属性(对于_1)(不太可能)或ruleStart_没有继承的属性(_r1)。

这就是我现在能告诉你的。

编辑是后者。ruleHandHeader不声明任何属性，更不用说要绑定到_r1的继承属性了

更新到评论。

以下是一些建议。与我经常重复的避免语义行为的建议(Boost Spirit："语义行为是邪恶的"？)非常相似，我会将结构调整为融合序列：

并使用简化得多的语法规则：

ruleStart_      = ruleEncoding_ >> ruleHandHeader_;
ruleEncoding_   = "ABC";
ruleHandId_     = qi::long_long;
ruleHandHeader_ = "DEF XYZ #" >> ruleHandId_;

现在，在规则定义中添加BOOST_SPIRIT_DEBUG宏并将uint64_t修复为uint64_t()：

住在科里鲁

#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/shared_ptr.hpp>
#include <sstream>
namespace qi = boost::spirit::qi;
struct parseContext {
std::string encoding_;
uint64_t    handId_;
};
BOOST_FUSION_ADAPT_STRUCT(parseContext, (std::string, encoding_)(uint64_t,handId_))
typedef boost::shared_ptr<parseContext> parseContextShPtr;
template <typename Iterator>
struct parseGrammar : qi::grammar<Iterator, parseContext()> {
parseGrammar() : parseGrammar::base_type(ruleStart_)
{
ruleStart_      = ruleEncoding_ >> ruleHandHeader_;
ruleEncoding_   = "ABC";
ruleHandId_     = qi::long_long;
ruleHandHeader_ = "DEF XYZ #" >> ruleHandId_;
BOOST_SPIRIT_DEBUG_NODES((ruleStart_)(ruleEncoding_)(ruleHandId_)(ruleHandHeader_))
}
// Rules
qi::rule<Iterator, parseContext()> ruleStart_;
qi::rule<Iterator, std::string()> ruleEncoding_;
qi::rule<Iterator, uint64_t()> ruleHandId_, ruleHandHeader_;
};
void test()
{
std::stringstream sb("ABCDEF XYZ #555: PQI #777");
sb.unsetf(std::ios::skipws);
typedef boost::spirit::istream_iterator It;
const parseGrammar<It> p;
It b(sb), e;
parseContextShPtr ctx(new parseContext);
bool r = qi::parse(b, e, p, *ctx);
if (r) {
std::cout << "Success: " << ctx->encoding_ << ", " << ctx->handId_ << std::endl;
}
else {
std::cout << "Failure" << std::endl;
}
if (b!=e)
std::cout << "Remaining: '" << std::string(b, e).substr(0, 32) << "'...n";
}
int main()
{
test();
}

指纹

Success: ABC, 555
Remaining: ': PQI #777'...
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