Thread: Convert a Julian Day to gregorian

I need to write a function in C that will be passed the current julian date for the current year for e.g

333 is today 28/11/2008 format required 20081128

60 is the 29/02/2008 format required 20080229

I have a function but it is very cumbersome and I am sure it could be far simpler than the way it is written.

Code:

char* getpaydate(char *date)
{
  time_t  currtime;
  char charJuli[8] = {0};
  char charYear[6] = {0};
  char caldate[8] = {0};
  int Yr,Dy,JDD;
  int q = 0;
  int Mt[] = {31,28,31,30,31,30,31,31,30,31,30,31};

  time(&currtime);
  strftime(charJuli,sizeof(charJuli)-1,"%j",localtime(&currtime));
  strftime(charYear,sizeof(charYear)-1,"%Y",localtime(&currtime));

  Yr = atoi(charYear);
  JDD = atoi(charJuli);
  Dy = atoi(date);

  if (Dy > 345 && JDD < 20)
  {
    Yr--;
  }
  if ((Yr/4)*4==Yr)
  {
    Mt[1] = 29;
  }
  if (((Yr/100)*100==Yr)&&((Yr/400)*400!=Yr))
  {
    Mt[1] = 28;
  }
  while ( Dy > Mt[q] )
  {
    Dy = Dy - Mt[q];
    q++;
  }
  q++;
  sprintf(caldate,"%i%02i%02i",Yr,q,Dy);
  printf("Start of the debug:%s\n", date );
  printf("Value of date equals :%s\n", date );
  printf("Value of charJuli equals :%s\n", charJuli );
  printf("Value of charYear equals :%s\n", charYear );
  printf("Value of caldate equals :%s\n", caldate );
  printf("Value of Yr equals :%d\n", Yr );
  printf("Value of JDD equals :%d\n", JDD    );
  printf("Value of Dy equals :%d\n", Dy );
  printf("End of Initial debug:%s\n", date );
  return caldate;
}

This if passed in a day of the year will get the current year and return the gregorian date 2008mmdd - so if it were returning today's date you would pass in the number 333 to get 20081128 and say 60 to get 20080229 (leap year).

Does this make sense ?

There was a super post of a function that presented the current julian date see below.

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_yda
y, jday);
         }
      }
   }
   return 0;
}

Thanks in advance.

The Julian date isn't always in the range of 1-366. OP is not working with Julian dates. For some reason people mix up the meaning of Julian dates and ordinal dates, when they are really not related at all.

The following function is based on a formula from Deshowitz and Reingold's Calendrical Calculations. Refer to to footnote #5 at Julian dates which is the same link from post #5. Now all you need to do is some repetitive subtraction from the ordinal date.

Code:

int DaysInMonth(int iYear, int iMonth) 
{
    return iMonth*275%9 > 3 ? 31
        : iMonth != 2     ? 30
        : iYear % 4     ? 28
        : iYear % 100  ? 29
        : iYear % 400 ? 28
        : 29;
}