Re: Question about caldate_mjd()

> Ooops .. let me restate that... Something, 111AD? (1859 years before 
> 1970 if I calculate right)  My confustion stems from this line:
> 
>   d = cd->day - 678882L;

I confess that I don't fully understand the algorithm, but I can shed
some light.

Modified Julian date 0 == 1858-11-17 00:00:00.

The 1970 epoch doesn't come into the caldate format; the `year' member
is simply the year (2003 CE => 2003, etc.).

So I'm reasonably sure that 678882 is the number of days from the
beginning of the common era till 1858-11-17.  (With a rough reckoning
I get to 678942 days, but I don't know enough historical background to
calculate the exact figure - if, indeed, there is one!)

Tim.