Re: [UUG/MLUG] perl and y2k

[David Lloyd <EMAIL:PROTECTED>]

I've also noticed on an accounting system I wrote that the POSIX::strftime
function expects the output from localtime.  So, if you add 1900 to the
year you get from localtime, you get something like the year 3900 in your
output.  So, if you use strftime make sure you add 1900 after the strftime
function call.  Alternately, you could use two variables, one with the
date as 100 and the other with the date as 2000 to fix your problem (damn,
now I've got to go back and change my code *again*).

Dave Lloyd

On Wed, 5 Jan 2000, Mark Donnelly wrote:

> >From http://www.perl.com/pub/1999/01/y2k.html, written by Tom
> Christiansen:
> 
> -=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
> The date and time functions supplied with Perl are the gmtime() and
> localtime()  functions, which are derived from their namesakes from the C
> programming language. These supply adequate information to determine the
> year well beyond 2000. 2038 is when trouble strikes, but only for those of
> us still stuck on 32-bit machines, a somewhat unlikely albeit admittedly
> not entirely unthinkable situation.
> 
>                     The year returned by these functions (when used in
> list context) is, contrary to popular misconception, not by definition a
> two-digit year. Rather, it merely happens to be such right now.  What it
> actually is, is the current year minus one thousand nine hundred. For
> years between 1900 and 1999 this happens to be a 2-digit decimal number,
> but that's not going to last long. To avoid the year 2000 problem, simply
> do not treat the year as a 2-digit number. Easy to say, and easy to break.
> Imagine that you want find out what the year appears to be in five years,
> so you write code like this.
> 
>                         
> use Time::localtime;
> $then = time() + ( 60 * 60 * 24 * 365 * 5 );  # 5 years from now
> $that_year = localtime($then) -> year;
> 
> printf("It shall be 19%d\n", $that_year); # WRONG! 19103
> printf("It shall be %d\n", 1900 + $that_year); # right:  2003
> 
> 
> 
>