Problems with INN-2.2.3 and SuSE6.4

[Tom Kacvinsky]

Oopsie!  My first stab didn't catch things like yyYear = 2001, $4 = 100.

Here is better take (with results below). Again, note the dependence on yyYear
being initialized to 1900 + (years since 1900).  yyIn corresponds to $4.

Tom

-- Cut here --

#include <stdlib.h>


typedef struct year_struct {
  int fouryear;
  int year;
} year;

int main () {

  int yyYear = 2000 ;
  int yyIn = 99 ;
  int i;

  year Years[] = {
    {2000,  97}, {2000,  98}, {2000,  99}, {2000, 100}, {2000, 101}, {2001,  99},
    {2001, 100}, {2001, 101}, {2002, 101}, {2002, 102}, {2098, 198}, {2099, 199}
  };

  for (i = 0; i < 12; i++ ) {
    yyIn = Years[i].year ;
    yyYear = Years[i].fouryear ;

    if (yyIn > 199) {
      yyYear = yyIn ;
    } else {

      if ( yyIn >= 100) {
        yyYear = 1900 + yyIn ;
      } else {
	yyYear = yyIn + (yyYear / 100 + (yyYear % 100 - yyIn) / 50) * 100 ;
      }
    }

    printf ("Four digit year: %d, Short Year: %d, Result: %d.\n", Years[i].fouryear, Years[i].year, yyYear) ;

  }

  return (0) ;

}

sun06> ./year
Four digit year: 2000, Short Year: 97, Result: 1997.
Four digit year: 2000, Short Year: 98, Result: 1998.
Four digit year: 2000, Short Year: 99, Result: 1999.
Four digit year: 2000, Short Year: 100, Result: 2000.
Four digit year: 2000, Short Year: 101, Result: 2001.
Four digit year: 2001, Short Year: 99, Result: 1999.
Four digit year: 2001, Short Year: 100, Result: 2000.
Four digit year: 2001, Short Year: 101, Result: 2001.
Four digit year: 2002, Short Year: 101, Result: 2001.
Four digit year: 2002, Short Year: 102, Result: 2002.
Four digit year: 2098, Short Year: 198, Result: 2098.
Four digit year: 2099, Short Year: 199, Result: 2099.




On Sun, 10 Dec 2000, Tom Kacvinsky wrote:

>
> Hi,
>
> When the code Russ quotes is invoked, yyYear is 2000 (yyYear is initialized to
> tm->tm_year + 1900 later in the code, which [on a Solaris box] is 1900 + 100 =
> 2000).  With $4 = 99, the code works and yyYear is set to 1999.  When $4 = 100,
> the year returned is 1900.  It looks like the code is broken in case $4 = 100.
> I think it is broken when $4 = 101, even.  Think about it.  As the code is
> written when $4 > 100, the year returned is 101.  Why should we worry about
> this?  Because the news clients that are writing out 100 for the year 2000 are
> most likely going to write out 101 for the year 2001.
>
> We probably need to consider 3 caes:
>
> case 1: 0 <= $4 < 100
>
> case 2: 100 <= $4 <= 199
>
> case 3: 199 < $4
>
> I chose 199 as the cutoff because by 2099, one would hope that clients no longer
> write out 20xx as 1xx, and 2099 is the last year of the form 20xx that can be
> erroneously written as 1xx.  I hope that makes sense!  I have been trying ot
> find a terse wording of this for the last 1/2 hour.  Blame the mathematician in
> me!
>
> Something like this:
>
>   if ($4 > 199) {
>     yyYear = $4;
>   } else {
>     yyYear = $4 + (yyYear / 100 + (yyYear % 100 - $4) / 50) * 100;
>     if ($4 >= 100)
>       yyYear = yyYear + 100;
>   }
>
> Below is some sample code, and the results.  Please note, I do not know if
> yyYear will always be initialized to 1900 + (numbers of years since 1900).  This
> is the weak link in the code below.
>
> Tom
>
> -- cut here --
> #include <stdlib.h>
>
>
> typedef struct year_struct {
>   int fouryear;
>   int year;
> } year;
>
> int main () {
>
>   int yyYear = 2000 ;
>   int yyIn = 99 ;
>   int i;
>
>   year Years[] = {{2000,97}, {2000,98}, {2000,99}, {2000,100}, {2000,101}, {2001,101}, {2002,102}, {2098,198}, {2099,199}};
>
>   for (i = 0; i < 9; i++ ) {
>     yyIn = Years[i].year ;
>     yyYear = Years[i].fouryear ;
>
>
>     if (yyIn > 199) {
>       /* assume year is YYYY format, so need not to add 1900 */
>       yyYear = yyIn ;
>     } else {
>
>       /* assume year is YY format, so need to add 1900 */
>       yyYear = yyIn + (yyYear / 100 + (yyYear % 100 - yyIn) / 50) * 100 ;
>
>       if ( yyIn >= 100)
> 	yyYear = yyYear + 100 ;
>
>     }
>
>     printf ("Four digit year: %d, Short Year: %d, Result: %d.\n", Years[i].fouryear, Years[i].year, yyYear) ;
>
>   }
>
>   return (0) ;
>
> }
>
> sun06> ./year
> Four digit year: 2000, Short Year: 97, Result: 1997.
> Four digit year: 2000, Short Year: 98, Result: 1998.
> Four digit year: 2000, Short Year: 99, Result: 1999.
> Four digit year: 2000, Short Year: 100, Result: 2000.
> Four digit year: 2000, Short Year: 101, Result: 2001.
> Four digit year: 2001, Short Year: 101, Result: 2001.
> Four digit year: 2002, Short Year: 102, Result: 2002.
> Four digit year: 2098, Short Year: 198, Result: 2098.
> Four digit year: 2099, Short Year: 199, Result: 2099.
>
> Tom
>
> >
> > Hm.  Looking at parsedate.y in 2.3, there are actually a whole bunch of
> > places that should be handling this:
> >
> >             if ($4 > 100) {
> >                 /* assume year is YYYY format, so need not to add 1900 */
> >                 yyYear = $4;
> >             } else {
> >                 /* assume year is YY format, so need to add 1900 */
> >                 yyYear = $4 + (yyYear / 100 + (yyYear % 100 - $4) / 50) * 100;
> >             }
> >
> > I'm not sure why it's not doing what you want.  Hurm.  Although I also
> > have no idea what that formula is really doing, since it's not obviously
> > just adding 1900.
> >
> >
>
>
>
>
>
>