Boost Spirit X3:将(一些）空格解析为枚举

我关于船长问题的参考：提升精神船长问题

在这种情况下，您使其与no_skip[]一起使用。没错。

no_skip就像lexeme一样，除了它没有从源代码(boost/spirit/home/x3/directive/no_skip.hpp(预先跳过：

// same as lexeme[], but does not pre-skip

另类拍摄

在您的情况下，我会颠倒逻辑：只需调整船长本身。

另外，不要向船长提供phrase_parse，因为您的语法对船长的正确值高度敏感。

您的整个语法可能是：

const auto p  = x3::skip(x3::space - whitespace) [
*(+x3::graph | whitespace)
];

这是关于Coliru的现场演示

#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <iostream>
#include <iomanip>
namespace x3 = boost::spirit::x3;
enum Whitespace { NEWLINE, TAB, PIPE };
struct whitespace_p : x3::symbols<Whitespace> {
whitespace_p() {
add
("n", Whitespace::NEWLINE)
("t", Whitespace::TAB)
("|", Whitespace::PIPE)
;
}
} static const whitespace;
struct Element : x3::variant<Whitespace, std::string> {
using base_type::base_type;
using base_type::operator=;
};
using Elements = std::vector<Element>;
static inline std::ostream& operator<<(std::ostream& os, Element const& el) {
struct print_visitor {
std::ostream& os;
auto& operator()(Whitespace ws) const {
switch(ws) {
case Whitespace::NEWLINE: return os << "[newline]";
case Whitespace::PIPE: return os << "[pipe]";
case Whitespace::TAB: return os << "[tab]";
}
return os << "?";
}
auto& operator()(const std::string& str) const { return os << std::quoted(str); }
} vis{os};
return boost::apply_visitor(vis, el);
}
int main() {
std::string const text = "tHello n World";
auto start = begin(text), stop = end(text);
const auto p  = x3::skip(x3::space - whitespace) [
*(+x3::graph | whitespace)
];
Elements elements;
if (!parse(start, stop, p, elements)) {
std::cout << "failed to parse!n";
} else {
std::copy(begin(elements), end(elements), std::ostream_iterator<Element>(std::cout, "n"));
}
if (start != stop) {
std::cout << "unparsed: " << std::quoted(std::string(start, stop)) << 'n';
}
}

指纹

[tab]
"Hello"
[newline]
"World"

甚至更简单？

这里似乎根本不需要任何船长。为什么不呢：

const auto p  = *(+~x3::char_("nt|") | whitespace);

当我们使用它时，不需要符号来映射枚举：

struct Element : x3::variant<char, std::string> {
// ...
};
using Elements = std::vector<Element>;

然后

const auto p
= x3::rule<struct ID, Element> {}
= +~x3::char_("nt|") | x3::char_;

住在科里鲁

#include <boost/spirit/home/x3.hpp>
#include <boost/spirit/home/x3/support/ast/variant.hpp>
#include <iostream>
#include <iomanip>
namespace x3 = boost::spirit::x3;
struct Element : x3::variant<char, std::string> {
using variant = x3::variant<char, std::string>;
using variant::variant;
using variant::operator=;
friend std::ostream& operator<<(std::ostream& os, Element const& el) {
struct print_visitor {
std::ostream& os;
auto& operator()(char ws) const {
switch(ws) {
case 'n': return os << "[newline]";
case 't': return os << "[pipe]";
case '|': return os << "[tab]";
}
return os << "?";
}
auto& operator()(const std::string& str) const { return os << std::quoted(str); }
} vis{os};
return boost::apply_visitor(vis, el);
}
};
using Elements = std::vector<Element>;
int main() {
std::string const text = "tHello n World";
auto start = begin(text);
auto const stop = end(text);
Elements elements;
const auto p
= x3::rule<struct ID, Element> {}
= +~x3::char_("nt|") | x3::char_;
if (!parse(start, stop, *p, elements)) {
std::cout << "failed to parse!n";
} else {
std::copy(begin(elements), end(elements), std::ostream_iterator<Element>(std::cout, "n"));
}
if (start != stop) {
std::cout << "unparsed: " << std::quoted(std::string(start, stop)) << 'n';
}
}

指纹

[pipe]
"Hello "
[newline]
" World"

问题是您在第 76 行使用了 phrase_parser 而不是解析器。 尝试使用类似的东西

bool result =
parse(start, stop, elementsParser, elements);

你的phrase_parser被指示跳过空格，你真正不想要的。

看看第一个答案 如何使用boost：：spirit将一系列单词解析为向量？