Member Avatar for animesh

how to subtract two dates to find no of days .

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@Dave Sinkula is on the right track using the C library time APIs. The big caveat with std::tm + mktime is that it works in local time and will account for daylight saving transitions. If your dates sit near a DST change, midnight-to-midnight may be 23 or 25 hours and a naive difftime(...)/(60*60*24) can be off by one. Also remember tm_mon is 0-based and tm_year is years since 1900. And : an enum for weekdays does not solve calendar date differences across months/years.

If you have C++20, prefer the calendrical types in <chrono>. They model civil dates directly and compute differences in whole civil days without timezone/DST surprises:

#include <chrono>
#include <iostream>

int main() {
    using namespace std;
    using namespace std::chrono;

    // June 24, 2004 to July 5, 2004
    sys_days a = year{2004}/month{6}/day{24};
    sys_days b = year{2004}/month{7}/day{5};

    days diff = b - a;                // exact count of civil days
    cout << diff.count() << " days\n"; // prints 11
}

Stuck on pre-C++20? You can still make the C APIs reliable:

  • Set a safe midday hour to avoid DST edges: tm_hour = 12; tm_isdst = -1;
  • Convert both dates consistently (both UTC via timegm if available, or both local via mktime) before subtracting.
  • Divide the seconds by 86400 and be explicit about inclusive vs exclusive ranges (add 1 if you mean to count both endpoints).

As @Ancient Dragon and noted, keep the solution simple and robust; avoid overly complex or brittle date code unless you truly need time-of-day or timezone handling.

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Member Avatar for meabed

i think that you can use

enum

it is as

structure

but enum list the elements by index as array .. for example

enum days_of_the_week ={sat,sun,mon,tue,wed,thu,fri };

there for .. sat has the value 0 and sun = 1 and mon = 2 and so on ..
so if u say mon - sat
the compiler wil supstract the values only ... ther for u will get 2 .

enum days_of_the_week ={sat,sun,mon,tue,wed,thu,fri };

there for .. sat has the value 0 and sun = 1 and mon = 2 and so on ..
so if u say mon - sat
the compiler wil supstract the values only ... ther for u will get 2 .

Edited by Dani because: Formatting fixed
Member Avatar for Dave Sinkula

This is one way.

[b]#include[/b] <iostream>
[b]#include[/b] <ctime> 
 
[b]int[/b] [b]main[/b]() 
{
[b]   struct[/b] std::tm a = {0,0,0,24,5,104}; /* June 24, 2004 */
[b]   struct[/b] std::tm b = {0,0,0,5,6,104}; /* July 5, 2004 */
   std::time_t x = std::[b]mktime[/b](&a);
   std::time_t y = std::[b]mktime[/b](&b);
[b]   if [/b]( x != (std::time_t)(-1) && y != (std::time_t)(-1) )
   {
	 [b]double[/b] difference = std::[b]difftime[/b](y, x) / (60 * 60 * 24);
	 std::[b]cout[/b] << std::ctime(&x);
	 std::[b]cout[/b] << std::ctime(&y);
	 std::[b]cout[/b] << "difference = " << difference << " days" << std::[b]endl[/b];
   }
[b]   return[/b] 0; 
}
 
/* my output
Thu Jun 24 01:00:00 2004
Mon Jul 05 01:00:00 2004
difference = 11 days
*/
Member Avatar for Ancient Dragon

you are only 8 years too late with your post, and that is a horribly complex program for just subtracting two dates.

Member Avatar for deceptikon

Not to mention it wouldn't pass code review in any development house I've worked with (assuming it compiled). The code itself is brittle and the solution is clearly written by someone who hasn't done much date and time programming.

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