Julian day of year

Folks,
I'm trying to program some code that will return me the julian day of year, that means the day from 1 to 365 (or 366 in leap years).

All I need the specific part of code to do is: Get TODAY's Julian day value and put it as a number, int variable would do.

Thanks for the help,

Show us what you got so far

have a gander at this

I guess I should have asked how to use the time.h functions... Actually, I have very few knowledge on those functions work.

Is there any way I can just command the code to get me the day of the current date and convert it to julian myself ???

Thanks and sorry for asking the question in a bad way...

After reading that, I can only feel myself a little more dumb.

So what I really need is the "day of year". Got a clue or have a nice tutorial I can dive in reading to learn the time.h funcions ???

Thanks

Code:

---

#include <stdio.h>
#include <time.h>

int main(void)
{
  time_t now;
  if ( time(&now) != (time_t)(-1) )
  {
      struct tm *mytime = localtime(&now);
      if ( mytime )
      {
        char jday [ 4 ];
        if ( strftime(jday, sizeof jday, "%j", mytime) )
        {
            printf("mytime->tm_yday = %d, jday = \"%s\"\n", mytime->tm_yday, jday);
        }
      }
  }
  return 0;
}

/* my output
mytime->tm_yday = 165, jday = "166"
*/

---

Thanks everyone for helping.

Dave, I was insipired by one post of yours in another thread and it worked. Thanks too!!

See ya