在 Boost::Spirit 中,我如何解析后跟分号或带有可选分号的换行符的条目?
示例输入,其中每个条目都是一个 int 和一个 double:
12 1.4;
63 13.2
2423 56.4 ; 5 8.1
下面是示例代码,仅解析后跟空格的条目:
#include <iostream>
#include <boost/foreach.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/support_istream_iterator.hpp>
#include <boost/fusion/include/std_pair.hpp>
namespace qi = boost::spirit::qi;
typedef std::pair<int, double> Entry;
template <typename Iterator, typename Skipper>
struct MyGrammar : qi::grammar<Iterator, std::vector<Entry>(), Skipper> {
MyGrammar() : MyGrammar::base_type(entries) {
entry = qi::int_ >> qi::double_;
entries = +entry;
}
qi::rule<Iterator, Entry(), Skipper> entry;
qi::rule<Iterator, std::vector<Entry>(), Skipper> entries;
};
int main() {
typedef boost::spirit::istream_iterator It;
std::cin.unsetf(std::ios::skipws);
It it(std::cin), end;
MyGrammar<It, qi::space_type> entry_grammar;
std::vector<Entry> entries;
if (qi::phrase_parse(it, end, entry_grammar, qi::space, entries)
&& it == end) {
BOOST_FOREACH(Entry const& entry, entries) {
std::cout << entry.first << " and " << entry.second << std::endl;
}
}
else {
std::cerr << "FAIL" << std::endl;
exit(1);
}
return 0;
}
现在,为了按照我想要的方式解析(每个条目后跟分号或带有可选分号的换行符),我替换了这个:
entries = +entry;
通过这个:
entries = +(entry >> (qi::no_skip[qi::eol] || ';'));
其中 boost::spirit
运算符 ||
表示:(a 后跟可选的 b)或 b。但是,如果在此示例输入中 1.4
之后有空格,则会出错:
12 1.4
63 13.2
由于 no_skip
,空间不匹配是有道理的,但我找不到解决方案。
最佳答案
这是我的看法。
- 您可能想了解
qi::blank
(除qi::eol
外,它是qi::space
)。这将消除对no_skip
的需要。 核心语法变为:
entry = qi::int_ >> qi::double_; entries = entry % +qi::char_("\n;") >> qi::omit[*qi::space];
使用 BOOST_SPIRIT_DEBUG 了解解析失败的位置和原因(例如回溯)
输出:
12 and 1.4
63 and 13.2
2423 and 56.4
5 and 8.1
完整代码:
//#define BOOST_SPIRIT_DEBUG
#include <iostream>
#include <boost/foreach.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/support_istream_iterator.hpp>
#include <boost/fusion/include/std_pair.hpp>
namespace qi = boost::spirit::qi;
typedef std::pair<int, double> Entry;
template <typename Iterator, typename Skipper>
struct MyGrammar : qi::grammar<Iterator, std::vector<Entry>(), Skipper> {
MyGrammar() : MyGrammar::base_type(entries) {
entry = qi::int_ >> qi::double_;
entries =
entry % +qi::char_("\n;") // the data
>> qi::omit[*qi::space] > qi::eoi; // trailing whitespace
BOOST_SPIRIT_DEBUG_NODE(entry);
BOOST_SPIRIT_DEBUG_NODE(entries);
}
qi::rule<Iterator, Entry(), Skipper> entry;
qi::rule<Iterator, std::vector<Entry>(), Skipper> entries;
};
int main() {
typedef boost::spirit::istream_iterator It;
std::cin.unsetf(std::ios::skipws);
It it(std::cin), end;
MyGrammar<It, qi::blank_type> entry_grammar;
std::vector<Entry> entries;
if (qi::phrase_parse(it, end, entry_grammar, qi::blank, entries)
&& it == end) {
BOOST_FOREACH(Entry const& entry, entries) {
std::cout << entry.first << " and " << entry.second << std::endl;
}
}
else {
std::cerr << "FAIL" << std::endl;
exit(1);
}
return 0;
}
关于c++ - 如何解析后跟分号或换行符的条目(boost::spirit)?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10669287/