在boost::spirit::traits::transform_attribute中指示解析失败的正确方法是什么？我可以抛弃任何旧的异常，还是有特殊的事情要它让我做？
namespace boost
{
namespace spirit
{
namespace traits
{
template <>
struct transform_attribute<TwoNums, std::vector<char>, qi::domain>
{
typedef std::vector<char> type;
static type pre(TwoWords&) { return{}; }
static void post(TwoWords& val, type const& attr) {
std::string stringed(attr.begin(), attr.end());
//https://stackoverflow.com/questions/236129/the-most-elegant-way-to-iterate-the-words-of-a-string
std::vector<std::string> strs;
boost::split(strs, stringed, ",");
if(strs.size()!=2)
{
//What do I do here?
}
val = TwoWords(strs[0],strs[1]);
}
static void fail(FDate&) { }
};
}
}
}
答案 0 :(得分：1)
是的，引发异常似乎是唯一的带外方法。
您可以使用qi::on_error来捕获并响应它。
但是，尚不清楚您需要什么。在解析器中使用split似乎有点颠倒。拆分基本上是糟糕的解析版本。
为什么没有规则进行子解析？
#include <boost/algorithm/string.hpp>
#include <boost/spirit/include/qi.hpp>
#include <iomanip>
namespace qi = boost::spirit::qi;
struct Invalid {};
struct TwoWords {
std::string one, two;
};
namespace boost { namespace spirit { namespace traits {
template <> struct transform_attribute<TwoWords, std::vector<char>, qi::domain> {
typedef std::vector<char> type;
static type pre(TwoWords &) { return {}; }
static void post(TwoWords &val, type const &attr) {
std::string stringed(attr.begin(), attr.end());
std::vector<std::string> strs;
boost::split(strs, stringed, boost::is_any_of(","));
if (strs.size() != 2) {
throw Invalid{};
}
val = TwoWords{ strs.at(0), strs.at(1) };
}
static void fail(TwoWords &) {}
};
} } }
template <typename It>
struct Demo1 : qi::grammar<It, TwoWords()> {
Demo1() : Demo1::base_type(start) {
start = qi::attr_cast<TwoWords>(+qi::char_);
}
private:
qi::rule<It, TwoWords()> start;
};
int main() {
Demo1<std::string::const_iterator> parser;
for (std::string const input : { ",", "a,b", "a,b,c" }) {
std::cout << "Parsing " << std::quoted(input) << " -> ";
TwoWords tw;
try {
if (parse(input.begin(), input.end(), parser, tw)) {
std::cout << std::quoted(tw.one) << ", " << std::quoted(tw.two) << "\n";
} else {
std::cout << "Failed\n";
}
} catch(Invalid) {
std::cout << "Input invalid\n";
}
}
}
打印
Parsing "," -> "", ""
Parsing "a,b" -> "a", "b"
Parsing "a,b,c" -> Input invalid
这有点，因为这将要求您抛出expectation_failure。
这不是最佳选择，因为它假设您知道解析器将被实例化的迭代器。
on_error设计用于expectation points
#include <boost/algorithm/string.hpp>
#include <boost/spirit/include/qi.hpp>
#include <iomanip>
namespace qi = boost::spirit::qi;
struct Invalid {};
struct TwoWords {
std::string one, two;
};
namespace boost { namespace spirit { namespace traits {
template <> struct transform_attribute<TwoWords, std::vector<char>, qi::domain> {
typedef std::vector<char> type;
static type pre(TwoWords &) { return {}; }
static void post(TwoWords &val, type const &attr) {
std::string stringed(attr.begin(), attr.end());
std::vector<std::string> strs;
boost::split(strs, stringed, boost::is_any_of(","));
if (strs.size() != 2) {
throw qi::expectation_failure<std::string::const_iterator>({}, {}, info("test"));
}
val = TwoWords{ strs.at(0), strs.at(1) };
}
static void fail(TwoWords &) {}
};
} } }
template <typename It>
struct Demo2 : qi::grammar<It, TwoWords()> {
Demo2() : Demo2::base_type(start) {
start = qi::attr_cast<TwoWords>(+qi::char_);
qi::on_error(start, [](auto&&...){});
// more verbose spelling:
// qi::on_error<qi::error_handler_result::fail> (start, [](auto&&...){[>no-op<]});
}
private:
qi::rule<It, TwoWords()> start;
};
int main() {
Demo2<std::string::const_iterator> parser;
for (std::string const input : { ",", "a,b", "a,b,c" }) {
std::cout << "Parsing " << std::quoted(input) << " -> ";
TwoWords tw;
try {
if (parse(input.begin(), input.end(), parser, tw)) {
std::cout << std::quoted(tw.one) << ", " << std::quoted(tw.two) << "\n";
} else {
std::cout << "Failed\n";
}
} catch(Invalid) {
std::cout << "Input invalid\n";
}
}
}
打印
Parsing "," -> "", ""
Parsing "a,b" -> "a", "b"
Parsing "a,b,c" -> Failed
让我们假设一个更有趣的语法，其中您有;的{{1}}分隔列表：
TwoWords我们解析为"foo,bar;a,b"
的向量：
TwoWords我们没有使用特征来“强制”属性，而是改编了结构并依靠自动属性传播：
using Word = std::string;
struct TwoWords { std::string one, two; };
using TwoWordses = std::vector<TwoWords>;
解析器模拟数据类型：
BOOST_FUSION_ADAPT_STRUCT(TwoWords, one, two)
完整测试为 Live On Coliru
template <typename It>
struct Demo3 : qi::grammar<It, TwoWordses()> {
Demo3() : Demo3::base_type(start) {
using namespace qi;
word = *(graph - ',' - ';');
twowords = word >> ',' >> word;
start = twowords % ';';
}
private:
qi::rule<It, Word()> word;
qi::rule<It, TwoWords()> twowords;
qi::rule<It, TwoWordses()> start;
};
打印
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <iomanip>
namespace qi = boost::spirit::qi;
using Word = std::string;
struct TwoWords { std::string one, two; };
using TwoWordses = std::vector<TwoWords>;
BOOST_FUSION_ADAPT_STRUCT(TwoWords, one, two);
template <typename It>
struct Demo3 : qi::grammar<It, TwoWordses()> {
Demo3() : Demo3::base_type(start) {
using namespace qi;
word = *(graph - ',' - ';');
twowords = word >> ',' >> word;
start = twowords % ';';
}
private:
qi::rule<It, Word()> word;
qi::rule<It, TwoWords()> twowords;
qi::rule<It, TwoWordses()> start;
};
int main() {
using It = std::string::const_iterator;
Demo3<It> parser;
for (std::string const input : {
",",
"foo,bar",
"foo,bar;qux,bax",
"foo,bar;qux,bax;err,;,ful",
// failing cases or cases with trailing input:
"",
"foo,bar;",
"foo,bar,qux",
})
{
std::cout << "Parsing " << std::quoted(input) << " ->\n";
TwoWordses tws;
It f = input.begin(), l = input.end();
if (parse(f, l, parser, tws)) {
for(auto& tw : tws) {
std::cout << " - " << std::quoted(tw.one) << ", " << std::quoted(tw.two) << "\n";
}
} else {
std::cout << "Failed\n";
}
if (f != l) {
std::cout << "Remaining unparsed input: " << std::quoted(std::string(f,l)) << "\n";
}
}
}