epoch Prime

[Dieter]

(01-05-2018 06:03 PM)pier4r Wrote:  Now -> epoch
Then one has to count the days until today, caring for leap years.
(this means caring also on the various lengths of different months)
Once one has the days. Days*86400 plus the seconds passed today.

You don't have to count anything, let alone handle different month lengths. There are well-known formulas that take all this into account. Simply calculate the Julian day number for two dates and you're done.

(01-05-2018 06:03 PM)pier4r Wrote:  The tedious (but crucial) part would be: leap years algorithm and months definition.

Not at all.
Maybe you want to look at this HP35s program to see how easy all this can be done.

(01-05-2018 06:03 PM)pier4r Wrote:  epoch-> date
take the epoch and check how many days it fits.
Then check how many of those days fit years (considering leap years).
Then check how the rest of the days (after one has the number of years) fits in months (considering the different length of months).
Then once one has the remaining days , while year and month are identified, find the day and the hour.

Waayyyy too complicated. ;-)

The key is the conversion between dates and Julian day numbers. And this can be done easily on the Prime since, as far as I know, it has the required commands to add days to a calendar date or determine the difference between two days.

Example (date taken from the Wikipedia page on Unix time): Between 1 Jan 1970 and 17 Sept 2004 there were 12678 days (can be calculated directly with the according Prime command). This times 86400 gives Unix time 1095379200. Add the remaining seconds of the day and you're done.

Dieter