Re: [boost] [xpressive] capture float?

23 Jul 2014

      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 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::asstd::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 
    #include 

    int main()
    {
        namespace xpr = boost::xpressive;

        std::map 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::asstd::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

Re: [boost] [xpressive] capture float?

Eric Niebler