Returning a parser from a function breaks it

[maliberty]

Below is a test case that demonstrates parsing a simple syntax. In one case I have a function returning a parser (called endcap) and in the other the parser is contained in a single function. The former fails while the later works. I can't sort out what the difference is and am hoping for some help.

I'm using Boost 1.76 with gcc 8.3.1.

// g++ --std=c++17 -o parser parser.cpp && ./parser
// gcc version 8.3.1 20190311 (Red Hat 8.3.1-3) (GCC)
// Boost 1.76
#include <boost/spirit/include/qi.hpp>
#include <string_view>

namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;

using boost::spirit::ascii::space_type;
using boost::spirit::qi::lit;

using ascii::space;

auto endcap() { return lit("ENDCAP") >> lit(";"); }

template <typename Iterator> bool parse1(Iterator first, Iterator last) {
  auto type_rule = (lit("CLASS") >> endcap());

  bool valid = qi::phrase_parse(first, last, type_rule, space);

  return valid && first == last;
}

////////// Same as above but the endcap parser is inlined into the caller
template <typename Iterator> bool parse2(Iterator first, Iterator last) {
  auto type_rule = (lit("CLASS") >> lit("ENDCAP") >> lit(";"));

  bool valid = qi::phrase_parse(first, last, type_rule, space);

  return valid && first == last;
}

int main() {
  std::string str("CLASS ENDCAP ;");
  std::string_view s(str);
  std::cout << parse1(s.begin(), s.end()) << '\n';
  std::cout << parse2(s.begin(), s.end()) << '\n';
}

when I run this I get:

0
1

Thanks in advance for the help!

[Kojoley]

Duplicate of #436

Combining operators capture parsers by reference. You need to use qi::copy. Fixed repro https://wandbox.org/permlink/nio16CExgMHtVZ0f

[maliberty]

Thanks very much for the quick help!