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date.lqr
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date.lbr
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weekday.c
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2011-01-28
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/* *** weekday.c *** */
/* */
/* IBM - PC microsoft "C" */
/* */
/* function to determine the day of the week a given gregorian date */
/* falls on. Returns a 0 if successful or a -1 if not. */
/* */
/* WARNING - day must be declared to be at least 10 characters or a */
/* memory overwrite will occure. */
/* */
/* Written by L. Cuthbertson, March 1983 */
/* */
/*********************************************************************/
/* */
int weekday(indate,day)
char indate[],day[];
{
static char days [7][10] = {'S','u','n','d','a','y','\0',' ',' ',' ',
'M','o','n','d','a','y','\0',' ',' ',' ',
'T','u','e','s','d','a','y','\0',' ',' ',
'W','e','d','n','e','s','d','a','y','\0',
'T','h','u','r','s','d','a','y','\0',' ',
'F','r','i','d','a','y','\0',' ',' ',' ',
'S','a','t','u','r','d','a','y','\0',' '};
int iday;
long julian,gtoj();
/* get julian date */
if ((julian = gtoj(indate)) == (-1)) return(-1);
/* calculate day of week */
iday = (julian-1) % 7;
/* move weekday into character string */
strcpy(day,days[iday]);
/* done */
return(0);
}