/* * $Id: //pentools/main/datemath/util.c#3 $ * * written by:  Stephen J. Friedl *              Software Consultant *              Tustin, California USA *              steve@unixwiz.net / www.unixwiz.net * *	These are various utility functions that are used throughout *	the datemath program. */#include <stdio.h>#include <assert.h>#include <stdarg.h>#include <stdlib.h>#include <string.h>#include <ctype.h>#include "defs.h"int	century19_year = 40;	/* 0..39 = 2000 / 40..99 = 1999 *//* * jultoyymm() * *	Convert the Julian date to YYMM.  Note that if the jdate *	provided is < 1900, we treat it as 1900.  Return is -1 on *	error. */int jultoyymm(jdate_t jdate){short	mdy[3];	if (rjulmdy(jdate, mdy) < 0)		/* convert to mdy[]	*/		return(-1);	if (mdy[YY] < 1900)			/* too early?		*/		mdy[YY] = 1900;	mdy[YY] %= 100;				/* strip off 19xx	*/	return (mdy[YY]*100) + mdy[MM];}/* * yymmtojul.c * *	This converts a YYMM date to a Julian date, which is returned. *	The "eflag" indicates whether the date returned should be the *	end of the month or the beginning.  This assumes that YY is in *	*this* century, and it dies on error. */jdate_t yymmtojul(int yymm, int eflag){short	mdy[3];jdate_t	jdate;int	rv;	mdy[MM] = yymm % 100;	mdy[YY] = yymm / 100;	mdy[DD] = eflag ? daysinmm(mdy[MM], mdy[YY]) : 1;	mdy[YY] += 1900;	if (rv = rmdyjul(mdy, &jdate), rv < 0)		die("bad date conversion in yymmtojul-(%d)", rv);	else		return(jdate);}/* * yymm_add() * *	Given a YYMM date (which must be valid), add the given *	number of months to it.  The number of months may be *	negative, and the new YYMM is returned. */int yymm_add(int yymm, int nmonths){register int	mm = yymm % 100,	/* month of the guy	*/		yy = yymm / 100;	/* year of the guy	*/	mm += nmonths;	while (mm < 1)			/* year got smaller?	*/		mm += 12, yy--;	while (mm > 12)			/* year got bigger?	*/		mm -= 12, yy++;	return(yy * 100 + mm);}/* * die() * *	Given a format and some args, print an error message and *	exit.  The program depends on the external variable ProgName, *	which should be set to argv[0].  The format string should not *	contain a newline, as one is appended for you.  All output *	goes to stderr. */void die(const char *format, ...){va_list	args;	va_start(args, format);	if (ProgName)		(void)fprintf(stderr, "%s: ", ProgName);	vfprintf(stderr, format, args);	fputc('\n', stderr);	va_end(args);	exit(EXIT_FAILURE);}/* * year_to_yyyy() * *	Given a year that could be in two-digit or four-digit *	format, figure out a four-digit year for it. We have a *	variable cutoff for 1900/2000. */int year_to_yyyy(int year){	if ( year < century19_year )		year += 2000;	else if ( year < 100 )		year += 1900;	return year;}/* * daysinmm() * *	Given a month (1-based) and a year, return the number *	of days in that month.  The year may be either YY or YYYY *	with the full prefix.  We take care to properly calculate *	the number of days in February according to the following *	rule: * *		if the year is divisible by for, except years *		divisible by 100, but including years divisible *		by 400. * *	If there is no leading YY, 1900 is assumed. */int daysinmm(int mm, int yy){static int mdays[12] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };	if (mm < 1 || mm > 12)			/* bogus dates?		*/		return(-1);	if (mm != 2)				/* not February?	*/		return(mdays[mm-1]);	yy = year_to_yyyy(yy);#define		YYMOD(n)	((yy % (n)) == 0)	return(28 + (YYMOD(4) && (YYMOD(400) || !YYMOD(100))));}/* * daysinyymm.c * *	Given a YYMM date, return the number of days in that month. */int daysinyymm(int yymm){	return(daysinmm(yymm % 100, yymm / 100));}/* * strlower() * *	Given a string, convert it to all lower case. */char *strlower(char *s){char	*s_save = s;	assert(s != 0);	for (; *s; s++)	{		if (isupper(*s))			*s = tolower(*s);	}	return s_save;}