How do you print a C++11 time_point?
(In this post I will omit std::chrono::
qualifications for clarity. I trust you know where they go.)
The reason your code example fails to compile is that there is a mismatch between the return type of system_clock::now()
and the type of variable you are trying to assign this to (time_point<system_clock, nanoseconds>
).
The documented return value of system_clock::now()
is system_clock::time_point
, which is a typedef for time_point<system_clock, system_clock::duration>
. system_clock::duration
is implementation-defined, with microseconds
and nanoseconds
being commonly used. It seems that your implementation uses microseconds
, so the return type of system_clock::now()
is time_point<system_clock, microseconds>
.
time_point
s with different durations are not implicitly convertible to one another, so you get a compiler error.
You can explicitly convert time points with different durations using time_point_cast
, so the following would compile on your system:
time_point<system_clock, nanoseconds> time_point;
time_point = time_point_cast<nanoseconds>(system_clock::now());
Notice the explicit template parameter to time_point_cast
is the target duration type, not the target time_point type. The clock types must match in a time_point_cast
, so specifying the entire time_point type (which is templated on both the clock type and the duration type) would be redundant.
Of course in your case, since you are just looking to print the time point, there is no need for it to be at any specific resolution, so you can just declare time_point
to be the same type as what system_clock::now()
returns to begin with. A simple way to do that is to use the system_clock::time_point
typedef:
system_clock::time_point time_point;
time_point = system_clock::now(); // no time_point_cast needed
Since this is C++11, you can also just use auto
:
auto time_point = system_clock::now();
Having solved this compiler error, the conversion to time_t
works just fine:
std::time_t now_c = std::chrono::system_clock::to_time_t(time_point);
and you can now use standard methods for displaying time_t
values, like std::ctime
or std::strftime
. (As Cassio Neri points out in a comment to your question, the more C++-y std::put_time
function is not yet supported by GCC).
This snippet might help you:
#include <iomanip>
#include <iostream>
#include <chrono>
#include <ctime>
template<typename Clock, typename Duration>
std::ostream &operator<<(std::ostream &stream,
const std::chrono::time_point<Clock, Duration> &time_point) {
const time_t time = Clock::to_time_t(time_point);
#if __GNUC__ > 4 || \
((__GNUC__ == 4) && __GNUC_MINOR__ > 8 && __GNUC_REVISION__ > 1)
// Maybe the put_time will be implemented later?
struct tm tm;
localtime_r(&time, &tm);
return stream << std::put_time(&tm, "%c"); // Print standard date&time
#else
char buffer[26];
ctime_r(&time, buffer);
buffer[24] = '\0'; // Removes the newline that is added
return stream << buffer;
#endif
}
int main() {
std::cout << std::chrono::system_clock::now() << std::endl;
// Wed May 22 14:17:03 2013
}
Updated answer for an old question:
For a std::chrono::time_point<std::chrono::system_clock, some-duration>
there is now a 3rd party libraries that give you much better control. For time_points based on other clocks, there is still no better solution than to just get the internal representation and print it out.
But for system_clock
, using this library, this is as easy as:
#include "date.h"
#include <iostream>
int
main()
{
using namespace date;
using namespace std::chrono;
std::cout << system_clock::now() << " UTC\n";
}
which just output for me:
2016-07-19 03:21:01.910626 UTC
which is the current UTC date and time to microsecond precision. If on your platform system_clock::time_point
has nanosecond precision, it will print out nanosecond precision for you.