Mac Forum / Programming / Perl / August 2004

Here's the relevant bit from the Time::Local perldoc:

     o   Years in the range 0..99 are interpreted as shorthand for
years in
           the rolling "current century," defined as 50 years on either
side
           of the current year.

So, this being 2004, 55 is interpreted as 1955, while 54 is interpreted
as 2054.

So, what's wrong with the year 2054? That's covered later in the
perldoc:

       Please note, however, that the range of dates that can be
actually be
       handled depends on the size of an integer (time_t) on a given
platform.
       Currently, this is 32 bits for most systems, yielding an
approximate
       range from Dec 1901 to Jan 2038.

This applies to the default Perl for Mac OS X, which uses 32-bit
integers. This can be easily verified:

sherm:~ sherm$ perl -e 'use Time::Local;print ( (localtime(timelocal(0,
0, 0, 1, 1, 36)))[6], "\n");'
5
sherm:~ sherm$ perl -e 'use Time::Local;print ( (localtime(timelocal(0,
0, 0, 1, 1, 37)))[6], "\n");'
0
sherm:~ sherm$ perl -e 'use Time::Local;print ( (localtime(timelocal(0,
0, 0, 1, 1, 38)))[6], "\n");'
Cannot handle date (0, 0, 0, 1, 1, 2038) at -e line 1

sherm--

As Sherm has pointed out, it's due to the two-digit year wrap around.
(Remember the so-called Millennium bug? That's what you're perpetuating
in the above code)

Your best bet for anything to do with dates and times is to use the
DateTime modules. They work on all platforms and comprehensively handle
any date and time you can throw at it (right up to $MAXINT-12-31). They
handle localization and know all current and historic DST rules back to
1972.

Cheers!
Rick Measham

(disclaimer: I have developed some of the DateTime modules)

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