这里是X3新手。两个问题:
<attributes>[[1, 9, 2, ., 1, 6, 8, ., 1, 1, 1, ., 1, 1, 1], [8, 0]]</attributes>,当我期望类似于<attributes>[[1, 9, 2, ., 1, 6, 8, ., 1, ., 1], [8, 0]]</attributes>的东西时x3::repeat(1)[x3::digit],但这似乎是错误的,可能会导致第一个问题的错误。(使用x3::repeat(1)[x3::digit]是因为我似乎不能只使用x3::位,因为它会导致规则崩溃?)#include <iostream>
#include <string>
#define BOOST_SPIRIT_X3_DEBUG
#include <boost/spirit/home/x3.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
namespace x3 = boost::spirit::x3;
namespace ast
{
struct ip_port
{
std::string host;
boost::optional<std::string> port;
};
}
BOOST_FUSION_ADAPT_STRUCT(ast::ip_port, host, port)
namespace parser
{
template <typename T> auto as = [](auto name, auto p) { return x3::rule<struct _, T> {name} = p; };
const auto dec_octet = as<std::string>("dec_octet",
(
x3::char_('2') >> x3::char_('5') >> x3::char_('0', '5')
| x3::char_('2') >> x3::char_('0', '4') >> x3::digit
| x3::char_('1') >> x3::digit >> x3::digit
| x3::char_('1', '9') >> x3::digit
| x3::repeat(1)[x3::digit] // awkward way to force sequence from single char, but can't use x3::digit
)
);
const auto ipv4address = as<std::string>("ipv4address",
dec_octet >> x3::char_('.') >> dec_octet >> x3::char_('.') >> dec_octet >> x3::char_('.') >> dec_octet
);
const auto ip = as<std::string>("host", ipv4address);
const auto port = as<std::string>("port", +x3::digit);
const auto ip_port = as<ast::ip_port>("ip_port", ip >> -((':') >> port));
}
template <typename T, typename Parser>
bool parse(const std::string& in, const Parser& p)
{
T parsed;
auto iter = in.begin();
auto end_iter = in.end();
bool res = x3::parse(iter, end_iter, p, parsed);
return res && (iter == end_iter);
}
int main()
{
std::cerr << std::boolalpha << parse<ast::ip_port>(std::string{"192.168.1.1:80"}, parser::ip_port) << '\n';
return EXIT_SUCCESS;
}调试输出:
<ip_port>
<try>192.168.1.1:80</try>
<host>
<try>192.168.1.1:80</try>
<ipv4address>
<try>192.168.1.1:80</try>
<dec_octet>
<try>192.168.1.1:80</try>
<success>.168.1.1:80</success>
<attributes>[1, 9, 2]</attributes>
</dec_octet>
<dec_octet>
<try>168.1.1:80</try>
<success>.1.1:80</success>
<attributes>[1, 6, 8]</attributes>
</dec_octet>
<dec_octet>
<try>1.1:80</try>
<success>.1:80</success>
<attributes>[1, 1, 1]</attributes>
</dec_octet>
<dec_octet>
<try>1:80</try>
<success>:80</success>
<attributes>[1, 1, 1]</attributes>
</dec_octet>
<success>:80</success>
<attributes>[1, 9, 2, ., 1, 6, 8, ., 1, 1, 1, ., 1, 1, 1]</attributes>
</ipv4address>
<success>:80</success>
<attributes>[1, 9, 2, ., 1, 6, 8, ., 1, 1, 1, ., 1, 1, 1]</attributes>
</host>
<port>
<try>80</try>
<success></success>
<attributes>[8, 0]</attributes>
</port>
<success></success>
<attributes>[[1, 9, 2, ., 1, 6, 8, ., 1, 1, 1, ., 1, 1, 1], [8, 0]]</attributes>
</ip_port>
true谢谢。
发布于 2021-02-11 16:37:02
Q. 1.为什么结果包含重复的" 1,1,1"s,例如:[1,9,2,1,6,8,1,1,1,1,1,1,1,8,0],当我期望这样的结果[1,9,2,.,1,6,8,.,1,.,1,8,0]时
从上一次人们遇到这样的陷阱开始已经7天了
这是一个由来已久的“容器属性不是原子的”陷阱:
您可以使用qi::hold对其进行纸张处理。或者你可以修改你的策略。
就像在这种情况下,我建议使用raw来获取底层的源序列。
Q. 2.别那么尴尬..。
中间的步骤是
const auto dec_octet = x3::raw [ x3::uint_parser<uint8_t>{} ];轰隆隆。使用X3是一个高级解析器生成器这一事实。不要做那些琐碎、容易出错的工作。事实上,你可以简单地
const x3::uint_parser<std::uint8_t> dec_octet{};这使得“紧张化”发展到了需要的地步:
const x3::uint_parser<std::uint8_t> dec_octet{};
const x3::uint_parser<std::uint16_t> port{};
const auto ipv4address = x3::raw [
dec_octet >> '.' >> dec_octet >> '.' >> dec_octet >> '.' >> dec_octet ];
const auto ip_port = as<ast::ip_port>("ip_port", ipv4address >> -(':' >> port));吸脂术后
请注意端口使用uint16_t,x3::eoi期望完全解析,删除显式规则/转换:
住在Coliru
#include <iostream>
#include <string>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>
#include <boost/optional/optional_io.hpp>
#include <boost/spirit/home/x3.hpp>
namespace x3 = boost::spirit::x3;
namespace ast {
struct ip_port {
std::string host;
boost::optional<uint16_t> port;
};
using boost::fusion::operator<<;
}
BOOST_FUSION_ADAPT_STRUCT(ast::ip_port, host, port)
namespace parser {
const x3::uint_parser<uint8_t> dec_octet {};
const x3::uint_parser<uint16_t> port {};
const auto ipv4address = x3::raw[dec_octet >> '.' >> dec_octet >> '.'
>> dec_octet >> '.' >> dec_octet];
const auto ip_port = ipv4address >> -(':' >> port) >> x3::eoi;
}
template <typename Parser, typename Attr>
static inline bool parse(std::string_view in, Parser const& p, Attr& result)
{
return x3::parse(in.begin(), in.end(), p, result);
}
auto parse_ipport(std::string_view in)
{
ast::ip_port result;
if (!parse(in, parser::ip_port, result))
throw std::invalid_argument("ipv4address");
return result;
}
int main()
{
for (auto input : { "192.168.1.1:80", "1.1.1.1", ":" }) {
std::cerr << parse_ipport(input) << std::endl;
}
}打印
(192.168.1.1 80)
(1.1.1.1 --)
terminate called after throwing an instance of 'std::invalid_argument'
what(): ipv4address
Aborted (core dumped)将代码简化得更多一些optional
(192.168.1.1 80)
(1.1.1.1 0)
terminate called after throwing an instance of 'std::invalid_argument'
what(): ipv4address开箱即用
注意:您的语法不匹配所有符合RFC的ip v4地址。例如。
127.1对127.0.0.1有效。0177.1或0x7f.1也是要么真的修复它,要么不要重新发明轮子,使用boost::asio::ip::address_v4::from_string甚至boost::asio::ip::address::from_string,并免费获得IPv6支持。
https://stackoverflow.com/questions/66153694
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