year duration algorithm in c++ - c++

Question

I'm making a program that needs the duration (in time_t) of a year.
In other ways, time_t of DD/MM/YYYY + duration = time_t of DD/MM/YYYY+1
So it may not always be 365 days (and 29/02/2012 will become 28/02/2013)
Here's the algorithm I came with :
if YEAR is leap than
if we are before the 29th feb' than return 365+1 days
else if we are the 29th feb' than return 365-1 days
else return 365 days
else if YEAR+1 is leap than
if we are before or the 28th feb' than return 365 days
else return 365+1 days
else return 365 days
Here, a day is 60 * 60 * 24 seconds
This algorithm seems to work. But I was wondering if there were another way to do this without all theses conditions and only 2 possible return values, or just some "trick" to optimize the thing.
I tried to increment tm_year from the struct tm like this :
// t is the input time_t
struct tm Tm (*localtime(&t));
if (Tm.tm_mon == 2 && Tm.tm_mday == 29) --Tm.tm_mday;
++Tm.tm_year;
return mktime(&Tm) - t;
But the result isn't what I want, I got -1 hour, or -25...
I guess it's because a year is not exactly 365 * 24 * 60 * 60.

Answer 1

I would use Boost for this, since it already implements what you are looking for:
#include <iostream>
#include <boost/date_time/gregorian/gregorian_types.hpp>
namespace date = boost::gregorian;
int main() {
date::date_period dp(date::date(2012, 6, 4), date::date(2013, 6, 4));
long days = dp.length().days();
std::cout << "Days between dates: " << days << std::endl;
}
If you wanted more precision, then you could use posix_time from Boost too:
namespace ptime = boost::posix_time;
...
ptime::ptime t1(date::date(2012, 6, 4), ptime::hours(0));
ptime::ptime t2(date::date(2013, 6, 4), ptime::hours(0));
ptime::time_duration td = t2 - t1;
std::cout << "Milliseconds: " << td.total_milliseconds() << std::endl;
Typically time_t is measured in seconds. Therefore, you would just need to call to td.total_seconds() to obtain the value that you are looking for.

Answer 2

if YEAR is leap than
if we are before the 29th feb' than return 365+1 days
else if we are the 29th feb' than return 365-1 days
else return 365 days
else if YEAR+1 is leap than
if we are before or the 28th feb' than return 365 days
else return 365+1 days
else return 365 days
simplifies to:
if (YEAR is leap)
if (< 29th Feb) return 365+1
if (= 29th Feb) return 365-1
else if (YEAR+1 is leap)
if (> 29th Feb) return 365+1
return 365
But why would you want to do this? It's much better to have readable code than "trick" optimizations.
As #betabandido has suggested, something like date(year+1, mon, day) - date(year, mon, day) would be much simpler, far more readable and able to handle leap years, leap seconds and September missing 11 days.

Answer 3

The length of a solar year is not a fixed number. The Gregorian calendar has invented a method to compensate for leap years which is not entirely precise. That goes "a year is a leap year if it is divisible by 4, unless it is divisible by 100, but is leap again if it is divisible by 400.
We in Iran have a more precise calendar, in which the years change the second Earth makes a full circle around the sun. In the same link you can see the average solar year as 365.2422 days and mean interval between spring equinoxes as 365.2424 days.
In this link more details are given about the length of the solar year (tropical year) in seconds.