Question

I am trying to write a function that takes in a parameter which is the offset days, and returns the date those many offset days from now. I can easily get the current date from below

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

int main() {
time_t t = time(0);   // get time now
struct tm * now = localtime( & t );
cout << (now->tm_year + 1900) << '-' 
     << (now->tm_mon + 1) << '-'
     <<  now->tm_mday
     << endl;
}

My question is if I change the now->tm_mday to now->tm_mday – offset, is it smart enough to do the month change or the year change as they may change.

Answer 1

No — (now->tm_year + 1900), (now->tm_mon + 1) and now->tm_mday are separate expressions and adding a new arithmetic operation to one will not affect the others.

Apply the offset to t instead, which is an integral value representing seconds since UNIX epoch. Then the change will carry through to the tm structure and, ultimately, each of your output expressions:

time_t t0 = time(0);               // now
time_t t1 = time(0) - 5;           // five seconds ago
time_t t2 = time(0) - 60*60*2;     // two hours ago
time_t t3 = time(0) - 60*60*24*5;  // five days ago

// (do try to avoid "magic numbers", though)