Adding #days to a date

**acidblue** · 09-13-2011

So far I have this bit of code for adding number of days to a date using the time.h module:

Code:

int main(int argc, char **argv)
{
	int numDays;
	printf ("Enter number of days to add ");
	scanf ("%d", &numDays);
	time_t now = time(NULL);
   
	struct tm* tm = localtime(&now);
   
	tm->tm_mday += numDays;
   
	cout << asctime(tm);  
	
	return 0;
}

This works ok until you add 30 or more days to the date, then it'll give me something like>>September 53 2011<< if I add 40 days .

What have I done wrong??

**Salem** · 09-13-2011

Use mktime to normalise the values, so Sept 31 would become Oct 1.

**Elysia** · 09-14-2011

This looks like C and not C++, so this thread is in the wrong section. For what reason did you put it here?

**acidblue** · 09-14-2011

Originally Posted by Salem

Use mktime to normalise the values, so Sept 31 would become Oct 1.

I've tried:

Code:

cout << mktime(tm);

But that gives the same results

**whiteflags** · 09-14-2011

Are you sure? mktime returns the day expressed as seconds from the epoch. You would have to take that result and make it into a string again with something like ctime().

**AndrewHunter** · 09-14-2011

umm.....

Code:

#include <iostream>
#include <ctime>

int main(void){

	tm *tmptr = NULL;
	time_t now = time(NULL);
	int numdays;

	tmptr = localtime(&now);

	std::cout << "Current date/time is: "<<asctime(tmptr)<<std::endl;
	std::cout << "Enter number of days to add: ";
	std::cin >> numdays;

	tmptr->tm_mday += numdays; //add number of days
	mktime(tmptr);//normalize time ptr

	std::cout << numdays << " days later will be: "<<asctime(tmptr)<<std::endl;
	
	return(0);
}

**acidblue** · 09-14-2011

Thanks AndrewHunter your example works.
Strange though all I changed was this:

Code:

struct tm* tm = localtime(&now);
   
	tm->tm_mday += numDays;
        mktime(tm);

To this:

Code:

tmptr = localtime(&now);
   
	tmptr->tm_mday += numDays;
	mktime(tmptr);

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