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    <p><code>mktime</code> can do almost all of the work for you. To find the number of days between two dates, each specified by a <code>struct tm</code>, you could use:</p> <pre><code>int days_between (struct tm *tsa, struct tm *tsb) { time_t a = mktime(tsa); time_t b = mktime(tsb); return (b - a)/(60*60*24); } </code></pre> <p>This divides the difference in seconds by the number of seconds in a day.</p> <p>So the days of the year could be calculated like this:</p> <pre><code>int days_of_year (int y) { struct tm tsa = {0}; struct tm tsb = {0}; tsa.tm_mday = 1; tsa.tm_year = y - 1900; tsb.tm_mday = 1; tsb.tm_year = y + 1 - 1900; return days_between(&amp;tsa, &amp;tsb); } </code></pre> <p>The days of the month is similar, but you would keep the year constant, and use the next month in <code>tsb</code>.</p> <p>To calculate the number of weekdays, there are always at least 8 in the first 28 days. Then, depending on how many days there are in the month, you have to consider the remaining days. The <code>tm_wday</code> field provides the days since Sunday for the date in the <code>struct tm</code>.</p> <pre><code> int dss = tsa.tm_wday; weekdays = 8; switch (days) { case 31: if (dss == 0 || dss == 6) weekdays += 1; dss = (dss + 1) % 7; case 30: if (dss == 0 || dss == 6) weekdays += 1; dss = (dss + 1) % 7; case 29: if (dss == 0 || dss == 6) weekdays += 1; default: break; } </code></pre> <p>The complete example can be found at <a href="http://ideone.com/qQo3U" rel="nofollow">http://ideone.com/qQo3U</a></p> <p>To calculate leap year, the logic for the Gregorian Calendar is: A year is a leap year if the year is divisible by 4, except if it is also divisible by 100, in which case it is a leap year if it is divisible by 400.</p> <pre><code>int is_leap_year (int y) { return (y % 4) ? 0 : ((y % 100) ? 1 : !(y % 400)); } </code></pre> <p>However, <code>mktime</code> already knows about this, and can figure it out for you, as has already been demonstrated.</p>
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