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POThe implementation of calculating the number of days between 2 dates
 

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    <p>What's the most efficient way to calculate the number of days between 2 dates? Basically I'm asking how our favourate datetime libraries are implemented.</p> <p>I quickly implemented a solution that is ~O(n) as I run through 1 iteration per 4 years. (Code attached below)</p> <p>I was asked by an intro to problem solving with computers class to implement this, but they're simply iterating through <em>everyday</em> instead of every 4 years.. so I'm not content with that solution and came up with the one below. However, is there a more efficient solution available? If so, how do they accomplish it?</p> <pre><code>#include &lt;iostream&gt; using namespace std; #define check_leap(year) ((year % 400 == 0) || ((year % 4 == 0) &amp;&amp; (year % 100 != 0))) #define debug(n) cout &lt;&lt; n &lt;&lt; endl int get_days(int month, bool leap){ if (month == 2){ if (leap) return 29; return 28; } else if (month == 1 || month == 3 || month == 5 || month == 7 || month == 8 || month == 10 || month == 12){ return 31; } else { return 30; } } int days[] = {31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334}; #define days_prior_to_month(n) days[n-2] int num_days_between(int month1, int day1, int month2, int day2, bool leap){ if (month2 &gt; month1) return ((days_prior_to_month(month2) - days_prior_to_month(month1+1)) + get_days(month1, leap) - day1 + 1 + day2) + ((leap &amp;&amp; month1 &lt;= 2 &amp;&amp; 2 &lt;= month2) ? 1 : 0); else if (month2 == month1) return day2; return -1; } int main(int argc, char * argv[]){ int year, month, day, year2, month2, day2; cout &lt;&lt; "Year: "; cin &gt;&gt; year; cout &lt;&lt; "Month: "; cin &gt;&gt; month; cout &lt;&lt; "Day: "; cin &gt;&gt; day; cout &lt;&lt; "Year 2: "; cin &gt;&gt; year2; cout &lt;&lt; "Month 2: "; cin &gt;&gt; month2; cout &lt;&lt; "Day 2: "; cin &gt;&gt; day2; int total = 0; if (year2 != year){ int leapyears = 0; total += num_days_between(month, day, 12, 31, check_leap(year)); debug(total); total += num_days_between(1, 1, month2, day2, check_leap(year2)); debug(total); int originalyear = year; year++; year = year + year % 4; while (year &lt;= year2-1){ leapyears += check_leap(year) ? 1 : 0; year += 4; } total += leapyears * 366; debug(total); total += max(year2 - originalyear - leapyears - 1, 0) * 365; debug(total); } else { total = num_days_between(month, day, month2, day2, check_leap(year)); } cout &lt;&lt; "Total Number of Days In Between: " &lt;&lt; total &lt;&lt; endl; system("PAUSE"); return 0; } </code></pre>
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