dealing with dates

I want to subtract date A from date B. Both are stored as mm/dd/yyyy.

Googling on datediff - as known in vb and subtract dates all with c++ hasn't yields much.

Any std or time.h functions or am I left to write my own?

The Boost Date-time library has functionality for this. You can literally subtract two date objects, using the minus operator.

http://www.boost.org/doc/html/date_time.html

You could put your yyyy, mm, and dd into a struct tm, use mktime to convert to a time_t, and then subtract two time_ts with difftime (if both dates are within the time's epoch). Something similar to this, perhaps.

Code:

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#include <iostream>
#include <time.h>

using namespace std;
int main()
{
    time_t current_time,past_time;
   
   
    time(&current_time);
   
    cout<<"Current time in time_t format is "<<current_time<<endl;
    cout<<"Current time in ctime format is "<<ctime(&current_time)<<endl;
   
    past_time = current_time -1000;
    cout<<"\nPast time in time_t format is "<<past_time<<endl;
    cout<<"Past time in ctime format is "<<ctime(&past_time)<<endl;
   
   
   
    cin.get();
    return 0;
}

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You can also use the code available in the FAQ for actually timing how long your program takes to run.

Thanks to all.

Dave - that code is perfect.

There is always the option of using Julian dates, as well:

Code:

---

#include <iostream>
#include <iomanip>
using namespace std;

#define GREGORIAN

const char *DAY_OF_WEEK[] = { "Monday", "Tuesday", "Wednesday",
  "Thursday", "Friday", "Saturday", "Sunday" };

 
void print_julian(double julian_date) {
  int Z = julian_date + 0.5;
  int W = (Z - 1867216.25) / 36524.25;
  int X = W / 4;
  int A = Z + 1 + W - X;
  int B = A + 1524;
  int C = (B - 122.1) / 365.25;
  int D = 365.25 * C;
  int E = (B - D) / 30.6001;
  int F = 30.6001 * E;

  int day_of_month = B - D - F;
  int month = E > 12 ? (E - 13) : (E - 1);//E - 1 or E-13 (must get number less than or equal to 12)
  int year = month <= 2 ? (C - 4715) : (C - 4716);//(if Month is January or February) or C-4716 (otherwise)

  cout << month << "/" << day_of_month << "/" << year << endl;
}

int main() {
  int month = 8,
      year = 2005,
      day = 23;
  double hour = 9,
      minute = 16,
      second = 30;

  int a = (14 - month) / 12;
  int y = year + 4800 - a;
  int m = month + (12 * a) - 3;
 
#ifdef GREGORIAN
  // for a date in the Gregorian calendar (at noon)
  int jdn = day +
    (((153 * m) + 2) / 5) +
    (365 * y) +
    (y / 4) -
    (y / 100) +
    (y / 400) -
    32045;
#else
  // for a date in the Julian calendar (at noon)
  int jdn = day + (((153 * m) + 2) / 5) + (365 * y) +
    (y / 4) - 32083;
#endif

  double julian_date = jdn + ((hour - 12) / 24) + (minute / 1440) + (second / 86400);
 
  cout << "The day of the week is " <<
    DAY_OF_WEEK[static_cast<int>(julian_date) % 7] << "." << endl;
 
  cout << "Today is ";
  print_julian(julian_date);
 
  julian_date -= 40;
  cout << "Forty days ago it was ";
  print_julian(julian_date);
 
  return 0;
}

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Quote:

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Originally Posted by the output of that program

The day of the week is Monday.
Today is 8/23/2005
Forty days ago it was 7/14/2005

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this code return this error :

12 E:\WKC\Untitled1.cpp [Warning] converting to `int' from `double' .

(if it is my mistake please teach me, im new to c++)

Quote:

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Originally Posted by wyvern

this code return this error :

12 E:\WKC\Untitled1.cpp [Warning] converting to `int' from `double' .

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It is a warning (not an error), and I think its message is pretty clear. difftime returns a double, which is being implicitly converted to an int -- thus drawing the warning.