Given an old-style const char * pointer and a length, is there a way to call std::regex_search() on it without first copying the contents of the buffer into a std::string? Here is a simple example of the problem I have:
#include <regex>
int main()
{
const char *text = "123 foobar 456";
const size_t len = strlen(text);
const std::regex rx(" (.+)bar");
std::smatch what;
std::regex_search( text, text+len, what, rx); // <- problematic line
return 0;
}
I thought the 5th std::regex_search() that takes two iterators is what I need, but I'm not fully understanding how to convert pointers to iterators. When I try to compile the code above, I get this:
g++ -std=c++11 test.cpp
test.cpp:11:45: error: no matching function for call to ‘regex_search(const char*&, const char*, std::smatch&, const regex&)’
/usr/include/c++/4.9/bits/regex.h:2131:5: note: template<class _Bi_iter, class _Alloc, class _Ch_type, class _Rx_traits> bool std::regex_search(_Bi_iter, _Bi_iter, std::match_results<_BiIter, _Alloc>&, const std::basic_regex<_CharT, _TraitsT>&, std::regex_constants::match_flag_type)
regex_search(_Bi_iter __s, _Bi_iter __e,
...and a lot more errors!
Can const char * be converted to the necessary iterator? Did I do it wrong? Am I misunderstanding how this works?