c++ Add or subtract values to the date - c++

Question

I couldn't figüre out how to add values to the date. Imagine, I have a date like 2012-5-15. After I parsed it, I obtained the year,month and day seperately and I tried to convert to my exact date to the days by using time_t and mktime. Do I have a chance to convert it back to the date from the days or seconds I found? Here is part of my code trying to do this;
if(operation="+"){
tm tm1=make_tm(year,mon,day);
time_t time1=mktime(&tm1);
int seconds_per_day=60*60*24;
time_t second=time1/seconds_per_day;
int sum=second //Trying to convert to the int from my time_t, don't know
//if it is neccesary
sum=sum+value //the value I want to add
So, basically if I give an input like 2012/5/15. It calculates how many days it passed by using the method but I just need to convert it to date and make it shown. Do you have any idea?

Answer 1

If you are asking for a way to add days to a particular date, here is a way to do it in similar terms to your attempt:
tm addDays(tm t1, int days)
{
time_t epoch = mktime(&t1);
epoch += (60*60*24 * days);
return *localtime(&epoch);
}
Then use the returned tm structure and access the tm_year, tm_mon and tm_mday members to get the yyyymmdd required output. A very rough and ready example (no proper formatting):
tm t1;
std::fill((char*)&t1, (char*)&t1 + sizeof(t1), 0);
t1.tm_year = 2016;
t1.tm_mon = 8;
t1.tm_mday = 2;
auto result = addDays(t1, 30);
std::cout << result.tm_year << '/' << result.tm_mon << '/' << result.tm_mday << std::endl;

Answer 2

Here is a free, open-source C++11/14 library to do what you want with very nice syntax, and unparalleled performance.
#include "date.h"
#include <iostream>
int
main()
{
using namespace date;
year_month_day x = sys_days{2012_y/may/15} + days{55};
std::cout << x << '\n';
}
This outputs:
2012-07-09
If you would prefer to write this yourself, here are the algorithms used by this date library. But the date library wraps everything up in a type-safe manner which catches many errors at compile time.
The date library is also nothing more than an extension of the <chrono> library, and so is very compatible with it:
#include "date.h"
#include <iostream>
int
main()
{
using namespace date;
using namespace std::chrono_literals;
auto x = sys_days{2012_y/may/15} + days{55} + 7h + 30min;
std::cout << x << '\n';
}
2012-07-09 07:30

Answer 3

thanks everybody for your attention but I figured out the problem in another way. Here is my way;
tm tm1=make_tm(year,mon,day);//of course I had an make_tm as struct
time_t time1=mktime(&tm1);
time1=time1+value*86400-86400 //because it was a second, I needed to convert the day to second. Then, I don't know why but it showed extra one day. So, again I subtracted 86400 then the problem solved.
char *cb=ctime(&time1);
cout<<cb<<endl;