[sylware]

Until you understand that the core of unix time is the "day", in the end, you only need to know the first leap year (If I recall properly it is 1972), then you have to handle the "rules" of leap years, and you will be ok (wikipedia I think, don't use google anymore since they now force javascript upon new web engines).

I did write such code in RISC-V assembly (for a custom command line on linux to output the statx syscall output). Then, don't be scared, with a bit of motivation, you'll figure it out.

[DamonHD]

The core of the UNIX time is seconds since epoch, nothing else. 'Day' has no special place at all. There are calendars for converting to and from dates, including Western-style, but the days in those calendars vary in length because of daylight saving switches and leap seconds for example.

[Kwpolska]

UNIX time ignores leap seconds, so every day is exactly 86400 seconds, and every year is either 365*86400 or 366*86400 seconds. This makes converting from yyyy-mm-dd to UNIX time quite easy, as you can just do `365*86400*(yyyy-1970) + leap_years*86400` to get to yyyy-01-01.

[sylware]

Yeap, this is why I said it is kind of easy.

Until you know properly the leap years. Leap year rules on the long run are are bit funky. Just have a look at wikipedia.

(do not use gogol search since they are now forcing javascript by default)

[XorNot]

Well yes, in the sense that not all Unix epoch seconds are equally long...

[sylware]

You are perfectly wrong, the day is the main calendar object related to the epoch seconds.

I wrote conversion code, I know what I am talking about.