The following code does compile (g++ 4.7.2
):
#include <chrono>
typedef std::chrono::duration< double > double_prec_seconds;
typedef std::chrono::time_point< std::chrono::system_clock > timepoint_t;
void do_something( const timepoint_t& tm )
{
// ...
}
int main( int argc, char** argv )
{
timepoint_t t0 = std::chrono::system_clock::now();
timepoint_t t1 = t0 + std::chrono::seconds(3);
// timepoint_t t3 = t0 + double_prec_seconds(3.14);
auto t3 = t0 + double_prec_seconds(3.14);
do_something( t1 );
}
My problem is that I don't know what type t3
has. It's not timepoint_t
, and un-commenting the line with the type explicitely given would not compile. The same with the function call: I can't call do_something
with t3
.
So my questions are:
I know that I can use an additional cast like this
// this works
timepoint_t t3 = t0 + std::chrono::duration_cast< std::chrono::milliseconds >(double_prec_seconds(3.14));
but I want to avoid this.
Thank you in advance!
The conversion fails, because there is no conversion from
std::chrono::time_point< std::chrono::system_clock,
std::chrono::system_clock::duration >
to
std::chrono::time_point< std::chrono::system_clock,
std::chrono::duration< double > >
The easiest way would be to give double_prec_seconds
as a template parameter to time_point
, see std::chrono::time_point
typedef std::chrono::time_point< std::chrono::system_clock,
double_prec_seconds > timepoint_t;
then you already have the proper type for t3
and do_something
.
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