In the example at http://en.cppreference.com/w/cpp/chrono the seconds value is obtained in a double
. This is the behavior I want. However, this appears to rely on the implicit assumption that the time point subtraction yields a value that represents seconds. I can not find anything that explains why a duration yields floating point seconds when no time units are actually specified. Without using auto
at all, can someone show how to explicitly obtain a time unit such as seconds in a floating point type, or point me to documentation or explanation as to why the aforementioned conversion represents seconds?
The conversion in question is essentially:
std::chrono::time_point<std::chrono::system_clock> start = std::chrono::system_clock::now();
std::chrono::time_point<std::chrono::system_clock> end = std::chrono::system_clock::now();
std::chrono::duration<double> elapsed_seconds = end-start; // why does this represent seconds as opposed to some other time unit?
double secondsInDouble = elapsed_seconds.count();
See the following posts for examples of the conversion.
How to get duration, as int milli's and float seconds from <chrono>?
C++ chrono - get duration as float or long long
According to this: http://en.cppreference.com/w/cpp/chrono/duration the default ratio for the second template parameter is 1:1
meaning seconds.
Other values are ratios relative to that. For example the ratio for std::chrono::milliseconds
is 1:1000
http://en.cppreference.com/w/cpp/numeric/ratio/ratio
So this statement:
std::chrono::duration<double> elapsed_seconds = end-start;
is equivalent to:
std::chrono::duration<double, std::ratio<1>> elapsed_seconds = end-start;
Which is defined to be seconds from which all other ratios are derived.
Whatever units end - start
are defined in get converted to std::ratio<1>
as in seconds.
If you wanted the time in milliseconds you could do:
std::chrono::duration<double, std::ratio<1, 1000>> elapsed_milliseconds = end-start;
And end - start
should be converted according to the new ratio.