long double future_time = static_cast(time(nullptr)) + (5.391247L * pow(10.0L, -44.0L))
Fixed-precision arithmetic would probably be more appropriate here, but since I’m lazy, long double works. Although I am curious now if a long double has sufficient precision to give a meaningful value for this.
t_furure_min = t_now + 5.391247 × 10^-44 seconds
long double future_time = static_cast(time(nullptr)) + (5.391247L * pow(10.0L, -44.0L))
Fixed-precision arithmetic would probably be more appropriate here, but since I’m lazy, long double works. Although I am curious now if a long double has sufficient precision to give a meaningful value for this.
I feel like at some point, you’d need arbitrary precision and arbitrarily-sized real numbers in order to meaningfully represent it.