Taking the first steps with <chrono> library, I'm starting with basic arithmetic on a days grained time_point. Thanks to a very useful post by @HowardHinnant, I managed to write this:

#include <chrono>
using namespace std::chrono_literals;

int main()
{
    std::chrono::sys_days d {std::chrono::January/31/2022};
    d  = std::chrono::days{2}; // ok
    //d  = 48h; // error: no match for 'operator =' with std::chrono::hours
}

What is not clear to me is why d = 48h; isn't allowed. The std::chrono::time_point<>::operator = takes a duration, and the rvalue in that expression is a std::chrono::hours that in my mind represents a time duration. What's the philosophy here? Are there different duration types according to the measure unit that must be compatible with the granularity of the time_point? Why?

On the other hand, I understand why d = 2d; gives an error, since in this case std::literals::chrono_literals::operator""d is a std::chrono::day, which is not a duration (that's handy to form a date literal, although it seems a little inconsistent to me). I wonder if there's a more convenient way to express a duration literal equivalent to std::chrono::days{2}.

CodePudding user response：

You can add hours to days. What you can't do is implicitly convert that into days again. You need a cast

d = std::chrono::time_point_cast<std::chrono::days>(d   48h);