C++ Cookbook

5.3. Performing Date and Time Arithmetic

Problem

You want to know the amount of time elapsed between two date/time points.

Solution

If both date/time points falls between the years of 1970 and 2038, you can use a time_t type and the difftime function from the <ctime> header. Example 5-6 shows how to compute the number of days elapsed between two dates.

Example 5-6. Date and time arithmetic with time_t

#include <ctime>
#include <iostream>
#include <cstdlib>

using namespace std;

time_t dateToTimeT(int month, int day, int year) {
  // january 5, 2000 is passed as (1, 5, 2000)
  tm tmp = tm();
  tmp.tm_mday = day;
  tmp.tm_mon = month - 1;
  tmp.tm_year = year - 1900;
  return mktime(&tmp);
}

time_t badTime() {
  return time_t(-1);
}

time_t now() {
  return time(0);
}

int main() {
  time_t date1 = dateToTimeT(1,1,2000);
  time_t date2 = dateToTimeT(1,1,2001);

  if ((date1 == badTime()) || (date2 == badTime())) {
    cerr << "unable to create a time_t struct" << endl;
    return EXIT_FAILURE;
  }
  double sec = difftime(date2, date1);
  long days = static_cast<long>(sec / (60 * 60 * 24));
  cout << "the number of days between Jan 1, 2000, and Jan 1, 2001, is ";
  cout << days << endl;
  return EXIT_SUCCESS;
}

The program in Example 5-6 should output :

the number of days between Jan 1, 2000, and Jan 1, 2001, is 366

Notice that the year 2000 is a leap year because even though it is divisible by 100; it is also divisible by 400, thus it has 366 days.

Discussion

The time_t type is an implementation defined arithmetic type. This means it ...