Date conversion algorithm - Help needed

[Se'noj]

Hi all,
In a program I am writing, I need to convert a date stored as an
integer number of days into a standard gregorian format, with day zero
being 1/1/1800. Can anyone help me with this?

Bryan Jones
NightDaemon Enterprises
nightdaemon.co. nr

[Victor Bazarov]

"Se'noj" <tl******@gmail .com> wrote...

In a program I am writing, I need to convert a date stored as an
integer number of days into a standard gregorian format, with day zero
being 1/1/1800. Can anyone help me with this?

So, what's really stopping you? Every year is 365 except the leap ones,
which are 366. Now, the number of days in your date, where is it counted
from?

[E. Robert Tisdale]

Se'noj wrote:

In a program I am writing,
I need to convert a date stored as an integer number of days
into a standard gregorian format,
with day zero being 1/1/1800.
Can anyone help me with this? cal 1 1800

January 1800
Su Mo Tu We Th Fr Sa
1 2 3 4
5 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31

I used Google

http://www.google.com/

to search for

+"C++" +"Gregorian date"

and I found lots of stuff.

[Micha]

Victor Bazarov wrote:

So, what's really stopping you? Every year is 365 except the leap ones,
which are 366. Now, the number of days in your date, where is it counted
from?

only nearly :-) every 100 hundred years the leap day is dropped, but there
are also exception (every 400 years I think), so 2000 was a leap year

[Victor Bazarov]

Micha wrote:

Victor Bazarov wrote:

So, what's really stopping you? Every year is 365 except the leap ones,
which are 366. Now, the number of days in your date, where is it counted
from?

only nearly :-) every 100 hundred years the leap day is dropped, but there
are also exception (every 400 years I think), so 2000 was a leap year

Perhaps you misunderstand what "leap" means, Micha? It means the year
has an extra day, thus making a leap, a jump, to catch up with the true
solar year. Did you misread what I wrote? I didn't say "every fourth
year has an extra day", did I? So, nothing "nearly" about what I wrote.

[Mike Wahler]

"Micha" <mi*****@fantas ymail.de> wrote in message
news:cm******** **@schlund.de.. .

Victor Bazarov wrote:

So, what's really stopping you? Every year is 365 except the leap ones,
which are 366. Now, the number of days in your date, where is it counted from?

only nearly :-) every 100 hundred years the leap day is dropped, but there
are also exception (every 400 years I think), so 2000 was a leap year

Yes, there are several rules for determining which years
are leap years, but this doesn't change the fact Victor
cited: leap years contain 366 days.

You can google for the calculation rules.
-Mike

[Micha]

Victor Bazarov wrote:

Perhaps you misunderstand what "leap" means, Micha? It means the year
has an extra day, thus making a leap, a jump, to catch up with the true
solar year. Did you misread what I wrote? I didn't say "every fourth
year has an extra day", did I? So, nothing "nearly" about what I wrote.

indeed my german is a little bit better than my english - although I know
what "leap" means (and I think the original poster also knows that there
are leap years)

but what I ment was, that even if I know that there are leap years, it isn't
obviously easy to catch all exceptions; so why invent such a boring
algorithm myself, if there is surely somebody out there who has already
done it with a nice little piece of code?

[Victor Bazarov]

Micha wrote:

[...]
but what I ment was, that even if I know that there are leap years, it isn't
obviously easy to catch all exceptions; so why invent such a boring
algorithm myself, if there is surely somebody out there who has already
done it with a nice little piece of code?

Let me get this straight. When you said (and I quote):

"only nearly :-) every 100 hundred years the leap day is dropped,
but there are also exception (every 400 years I think), so 2000
was a leap year"

you actually meant "don't reinvent the wheel"? Fascinating!

[osmium]

"Victor Bazarov" wrote:

[...]
but what I ment was, that even if I know that there are leap years, it
isn't
obviously easy to catch all exceptions; so why invent such a boring
algorithm myself, if there is surely somebody out there who has already
done it with a nice little piece of code?

Let me get this straight. When you said (and I quote):

"only nearly :-) every 100 hundred years the leap day is dropped,
but there are also exception (every 400 years I think), so 2000
was a leap year"

you actually meant "don't reinvent the wheel"? Fascinating!

LOL

[Marcelo Pinto]

"Mike Wahler" <mk******@mkwah ler.net> wrote in message news:<sh******* **********@news read3.news.pas. earthlink.net>. ..

"Micha" <mi*****@fantas ymail.de> wrote in message
news:cm******** **@schlund.de.. .

Victor Bazarov wrote:

So, what's really stopping you? Every year is 365 except the leap ones,
which are 366. Now, the number of days in your date, where is it counted from?
only nearly :-) every 100 hundred years the leap day is dropped, but there
are also exception (every 400 years I think), so 2000 was a leap year

Yes, there are several rules for determining which years
are leap years, but this doesn't change the fact Victor
cited: leap years contain 366 days.

You can google for the calculation rules.

int year;
//...
bool leap = ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0);


-Mike

--
Marcelo Pinto