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c++ calculating seconds in a year

Posted on 2011-03-08
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Last Modified: 2012-05-11
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;
}
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Question by:Gator77
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9 Comments
 
LVL 16

Expert Comment

by:sjklein42
ID: 35075158
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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Accepted Solution

by:
jkr earned 500 total points
ID: 35075204
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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LVL 11

Expert Comment

by:Kusala Wijayasena
ID: 35080226
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);
}

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-Kusala
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LVL 34

Expert Comment

by:sarabande
ID: 35081589
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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Expert Comment

by:sjklein42
ID: 35081848
Don't forget the occasional Leap-Second:

http://en.wikipedia.org/wiki/Leap_second
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Expert Comment

by:pjasnos
ID: 35090449
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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Expert Comment

by:evilrix
ID: 36518453
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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