On 7/23/2014 3:47 PM, Greg Rubino wrote:
On Wed, Jul 23, 2014 at 4:07 PM, Greg Rubino wrote:
Hi all,
I'm trying to capture a float with xpressive and put it into a local map. As a simplified example, I tried just pushing the matched pattern to a float and the result is always a float containing '0'.
namespace xpr = boost::xpressive;
std::map<float, std::string> l_floatOperMap;
xpr::sregex l_relation = xpr::as_xpr("<=") | ">=" | "<" | ">"); l_floatRestriction = !(xpr::s1 = l_relation) >> (xpr::s2 = (+xpr::_d >> '.' >> +xpr::_d)) [xpr::ref(l_floatOperMap)[xpr::as<float>(xpr::s2)] = xpr::as<std::string>(xpr::s1)];
However, when I go to print the contents of l_floatOperMap, it contains only the tuple (0, ""). Is it something I'm doing wrong? It seems like this should work to me.
I neglected to mention that I am actually doing the very same thing with uint32_t and it works fine.
Works for me. This program: #include <map> #include <iostream> #include <boost/xpressive/xpressive.hpp> #include <boost/xpressive/regex_actions.hpp> int main() { namespace xpr = boost::xpressive; std::map<float, std::string> l_floatOperMap; xpr::sregex l_relation = xpr::as_xpr("<=") | ">=" | "<" | ">"; xpr::sregex l_floatRestriction = !(xpr::s1 = l_relation) >> (xpr::s2 = (+xpr::_d >> '.' >> +xpr::_d)) [ xpr::ref(l_floatOperMap)[xpr::as<float>(xpr::s2)] = xpr::as<std::string>(xpr::s1) ]; std::string str = ">3.14"; if(xpr::regex_match(str, l_floatRestriction)) { for(auto p : l_floatOperMap) { std::cout << "(" << p.first << ", " << p.second << ")" << std::endl; } } } Gives me this expected output: (3.14, >) Maybe you could send a complete repro? Thanks, Eric