Right now I'm processing the following two regex replacements and I'm combining the substitution like this. Is there a way to process the replacement calling regex_replace one time only?
std::string test_string = "whatever";
std::regex reg_num("\\\"([0-9]+\\.{0,1}[0-9]*)\\\"");
std::regex reg_newline("\n+");
return std::regex_replace(std::regex_replace(test_string, reg_num, "$1"), reg_newline, "");
You can use
std::string test_string = "\"1\" \"1.50\"\nNew line";
std::regex reg_combined("\"([0-9]+\\.?[0-9]*)\"|\n");
std::cout << std::regex_replace(test_string, reg_combined, "$1");
See the C++ demo.
The $1 backreference is initialized with an empty string if Group 1 did not participate in the match, so it is safe to remove line feeds like this.
Pattern details:
" - a " char (note you do not need to escape it)([0-9]+\.?[0-9]*) - Group 1: one or more digits, an optional . and then zero or more digits (are you sure you do not need [0-9]*\.?[0-9]+?)" - a " char| - or\n - a line feed char.