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c++ calculating seconds in a year

how can i write a c++ program to determine  the number of seconds in any given year?

using namespace std;

int main(int argc, char *argv[])
{
    int year;
        cin >> year;
         cout << "leap year " << endl;
        }
        else
        {
        cout << " not leap year "  << endl;
    system("PAUSE");
    return EXIT_SUCCESS;
}
0
Gator77
Asked:
Gator77
1 Solution
 
sjklein42Commented:
I would do it by using two DateTime objects, one set to the time at the beginning of the year in question and the other set to the beginning of the next year.  Convert both DateTimes to seconds and subtract the result to get the length of the year in seconds.

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jkrCommented:
Use 'mktime()' (http://www.cplusplus.com/reference/clibrary/ctime/mktime/) for that, it returns a 'time_t' value for the date you pass in. 'time_t' now is nothing else but the amount of seconds elapsed since midnight (00:00:00), January 1, 1970 for any given date. Calculate that for Jan 1, Midnight and Dec 31 29:59 of that give year and subtract the results, then you have the amount of seconds in that year. Sorry for not being more specific, but EE has quite a strict homework policy.
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Kusala WijayasenaSoftware EngineerCommented:
If your intention is to calculate leap year, I would recommend following logic rather than something based on number of seconds in a year

bool isLeapYear(int year)
{
	return (year % 4 == 0) && (year % 100 != 0 || year % 400 == 0);
}

Open in new window



-Kusala
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sarabandeCommented:
obviously the number of seconds per year calculates by 366 * 24 * 3600 for leap years and 365 * 24 * 3600 else.

leap years we have for years dividable by 4 without remainder but not century years like 1900 beside the year is dividable by 400 without remainder like in 2000.

you can see the post of kusala for implementation of that rule.

Sara
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sjklein42Commented:
Don't forget the occasional Leap-Second:

http://en.wikipedia.org/wiki/Leap_second
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pjasnosCommented:
Leap seconds is definitely an overkill here - they can't be predicted algorithmically, so all you can do is include a big fat table of past values.
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evilrixSenior Software Engineer (Avast)Commented:
This question has been classified as abandoned and is closed as part of the Cleanup Program. See the recommendation for more details.
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