LibJS: Adjust approximated result in year_from_time() if necessary

This commit is contained in:
Linus Groh 2021-11-05 01:18:27 +01:00
parent 37146e305a
commit 1dc32fed8f
2 changed files with 19 additions and 1 deletions

View file

@ -201,11 +201,28 @@ double day_from_year(i32 y)
return 365 * (y - 1970) + floor((y - 1969) / 4.0) - floor((y - 1901) / 100.0) + floor((y - 1601) / 400.0);
}
// TimeFromYear(y), https://tc39.es/ecma262/#eqn-TimeFromYear
double time_from_year(i32 y)
{
// msPerDay × DayFromYear(y)
return MS_PER_DAY * day_from_year(y);
}
// YearFromTime(t), https://tc39.es/ecma262/#eqn-YearFromTime
i32 year_from_time(double t)
{
// the largest integral Number y (closest to +∞) such that TimeFromYear(y) ≤ t
return static_cast<i32>(t / (365.2425 * MS_PER_DAY) + 1970);
// Approximation using average number of milliseconds per year. We might have to adjust this guess afterwards.
auto year = static_cast<i32>(t / (365.2425 * MS_PER_DAY) + 1970);
auto year_t = time_from_year(year);
if (year_t > t)
year--;
else if (year_t + days_in_year(year) * MS_PER_DAY <= t)
year++;
return year;
}
// InLeapYear(t), https://tc39.es/ecma262/#eqn-InLeapYear

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@ -90,6 +90,7 @@ u16 day_within_year(double);
u8 date_from_time(double);
u16 days_in_year(i32);
double day_from_year(i32);
double time_from_year(i32);
i32 year_from_time(double);
bool in_leap_year(double);
u8 month_from_time(double);