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?
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.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
You can also use the code available in the FAQ for actually timing how long your program takes to run.Code:#include <iostream> #include <time.h> using namespace std; int main() { time_t current_time,past_time; time(¤t_time); cout<<"Current time in time_t format is "<<current_time<<endl; cout<<"Current time in ctime format is "<<ctime(¤t_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; }
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; }Originally Posted by the output of that program
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++)
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.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
