Reputation: 379
I'm getting myself familiarized with boost spirit v3. The question I want to ask is how to state the fact that you don't want to use skip parser in any way.
Consider a simple example of parsing comma-separated sequence of integers:
#include <iostream>
#include <string>
#include <vector>
#include <boost/spirit/home/x3.hpp>
int main()
{
using namespace boost::spirit::x3;
const std::string input{"2,4,5"};
const auto parser = int_ % ',';
std::vector<int> numbers;
auto start = input.cbegin();
auto r = phrase_parse(start, input.end(), parser, space, numbers);
if(r && start == input.cend())
{
// success
for(const auto &item: numbers)
std::cout << item << std::endl;
return 0;
}
std::cerr << "Input was not parsed successfully" << std::endl;
return 1;
}
This works totally fine. However, I would like to forbid having spaces in between (i.e. "2, 4,5"
should not be parsed well).
I tried using eps
as a skip parser in phrase_parse
, but as you can guess, the program ended up in the infinite loop because eps
matches to an empty string.
Solution I found is to use no_skip
directive (https://www.boost.org/doc/libs/1_75_0/libs/spirit/doc/html/spirit/qi/reference/directive/no_skip.html). So the parser now becomes:
const auto parser = no_skip[int_ % ','];
This works fine, but I don't find it to be an elegant solution (especially providing "space"
parser in phrase_parse
when I want no whitespace skips). Are there no skip parsers that would simply do nothing? Am I missing something?
Thanks for Your time. Looking forward to any replies.
Upvotes: 2
Views: 641
Reputation: 392833
You can use either no_skip[]
or lexeme[]
. They're almost identical, except for pre-skip (Boost Spirit lexeme vs no_skip).
Are there no skip parsers that would simply do nothing? Am I missing something?
A wild guess, but you might be missing the parse
API that doesn't accept a skipper in the first place
#include <iostream>
#include <iomanip>
#include <boost/spirit/home/x3.hpp>
namespace x3 = boost::spirit::x3;
int main() {
std::string const input{ "2,4,5" };
auto f = begin(input), l = end(input);
const auto parser = x3::int_ % ',';
std::vector<int> numbers;
auto r = parse(f, l, parser, numbers);
if (r) {
// success
for (const auto& item : numbers)
std::cout << item << std::endl;
} else {
std::cerr << "Input was not parsed successfully" << std::endl;
return 1;
}
if (f!=l) {
std::cout << "Remaining input " << std::quoted(std::string(f,l)) << "\n";
return 2;
}
}
Prints
2
4
5
Upvotes: 1