rom/utility/date2amiga.c

   1 /*
   2     Copyright © 1995-2001, The AROS Development Team. All rights reserved.
   3     $Id$
   4
   5     Desc: Convert a human understandable date to a machine form.
   6     Lang: english
   7 */
   8 #include "intern.h"
   9
  10 /*****************************************************************************
  11
  12     NAME */
  13 #include <utility/date.h>
  14 #include <proto/utility.h>
  15
  16         AROS_LH1(ULONG, Date2Amiga,
  17
  18 /*  SYNOPSIS */
  19         AROS_LHA(struct ClockData *, date, A0),
  20
  21 /*  LOCATION */
  22         struct UtilityBase *, UtilityBase, 21, Utility)
  23
  24 /*  FUNCTION
  25         Converts the information given in the struct ClockData *date, into
  26         the number of seconds that have past since the 1st of January 1978.
  27
  28     INPUTS
  29         date    -   Contains the information about the time.
  30
  31     RESULT
  32         The number of seconds since 1.1.1978
  33
  34     NOTES
  35
  36     EXAMPLE
  37
  38     BUGS
  39
  40     SEE ALSO
  41         Amiga2Date(), CheckDate()
  42
  43     INTERNALS
  44         Bit of a hack in the leap year handling.
  45
  46     HISTORY
  47         29-10-95    digulla automatically created from
  48                             utility_lib.fd and clib/utility_protos.h
  49
  50 *****************************************************************************/
  51 {
  52     AROS_LIBFUNC_INIT
  53
  54     /* This array contains the number of days that have been in the year
  55        up to the start of the month. Does not take into account leap years.
  56     */
  57     static const UWORD dayspermonth[12] = { 0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334 };
  58
  59     ULONG time;
  60     UWORD year, leaps;
  61
  62     year = date->year - 1978;
  63
  64     time = date->sec + (date->min * 60) + (date->hour * 3600);
  65     time += ((date->mday - 1) + dayspermonth[date->month - 1]) * 86400;
  66     time += year * 31536000;
  67
  68     /* How do we calculate leap years? That is a very good question.
  69        The year 1978 is NOT a leap year, but 1980 is...
  70
  71        I want to arrange it so that the last year of any four year group
  72        is a leap year. So the first year I can really deal with is either
  73        1977 or 1981. So I choose 1977 (the year I was born in :)
  74
  75         Then, to get the number of leap years I divide by 4.
  76         However, if year is a year which is a leap year, then this year
  77         will not be counted, so I have to check for this.
  78
  79         If year % 4 == 3, then we are in a leap year, and if the month
  80         is after February, then I can add a leap year.
  81     */
  82
  83 #if 1
  84     /* stegerg: based on dos.library/strtodate */
  85
  86     year = date->year;
  87
  88     if (((year % 400) == 0) || (((year % 4) == 0) && !((year % 100) == 0)))
  89     {
  90         /* data->year is a leap year. So dayspermonth table was wrong if
  91            month >= March */
  92
  93         if (date->month >= 3) time += 86400;
  94     }
  95
  96     year--;
  97
  98     leaps = ((year / 4) - (year / 100) + (year / 400) - (494 - 19 + 4));
  99
 100     time += leaps * 86400;
 101
 102 #else
 103     year++;
 104     leaps = year / 4;
 105     if( (year % 4 == 3) && (date->month > 2))
 106         leaps++;
 107
 108     /* If the year is greater than the year 2100, or it is the
 109        year 2100 and after February, then we also have to subtract
 110        a leap year, as the year 2100 is NOT a leap year.
 111     */
 112     if( ( year > 123) || ((year == 123) && (date->month > 2)) )
 113         leaps--;
 114
 115     time += leaps * 86400;
 116 #endif
 117
 118     return time;
 119
 120     AROS_LIBFUNC_EXIT
 121 } /* Date2Amiga */