If I do date +%H-%M-%S
on the commandline (Debian/Lenny), I get a user-friendly (not UTC, not DST-less, the time a normal person has on their wristwatch) time printed.
What's the simplest way to obtain the same thing with boost::date_time
?
If I do this:
std::ostringstream msg;
boost::local_time::local_date_time t =
boost::local_time::local_sec_clock::local_time(
boost::local_time::time_zone_ptr()
);
boost::local_time::local_time_facet* lf(
new boost::local_time::local_time_facet("%H-%M-%S")
);
msg.imbue(std::locale(msg.getloc(),lf));
msg << t;
Then msg.str()
is an hour earlier than the time I want to see. I'm not sure whether this is because it's showing UTC or local timezone time without a DST correction (I'm in the UK).
What's the simplest way to modify the above to yield the DST corrected local timezone time ? I have an idea it involves boost::date_time:: c_local_adjustor
but can't figure it out from the examples.
This does what I want:
namespace pt = boost::posix_time;
std::ostringstream msg;
const pt::ptime now = pt::second_clock::local_time();
pt::time_facet*const f = new pt::time_facet("%H-%M-%S");
msg.imbue(std::locale(msg.getloc(),f));
msg << now;
While this is not using boost::date_time it's relatively easy with boost::locale, which is quite more adapted for this task. As your need is simply getting a formatted time from the current locale.
IMHO boost::date_time should be used when you deal with softwares like gantt/planning computations, were you have alot of date_time arithmetic. But simply for using time and doing some arithmetic on it, you will faster success with boost::locale.
#include <iostream>
#include <boost/locale.hpp>
using namespace boost;
int main(int argc, char **argv) {
locale::generator gen;
std::locale::global(gen(""));
locale::date_time now;
std::cout.imbue(std::locale());
std::cout << locale::as::ftime("%H-%M-%S") << now << std::endl;
return 0;
}
Right now it should output : 15-45-48. :)
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