This drove me nuts for several hours today until I figured it out. Sample code below:

#!/usr/bin/perl
use strict;
use Time::Local;
#use bignum(p => -3);

my $key='121005';
print $key." ==> ".convtime($key)."\n";

sub convtime {
 my @months=qw(Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec);
 my @dow=qw{Sun Mon Tue Wed Thu Fri Sat};
 my $date=@_[0];
 my $year=substr $date,0,2;
 $year += 100;
 my $month=substr $date,2,2;
 $month -= 1;
 my $day=substr $date,4;
 my $gmt = timegm(0,0,0,$day,$month,$year);
 my @gmtime = gmtime($gmt);
 my $convstring=$dow[$gmtime[6]]." ".$months[$gmtime[4]]." ".$gmtime[3
+];
 return $convstring
}

[download]

If you run this, it returns "121005 ==> Fri Oct 5", as it should. If you uncomment the "use bignum" line, however, it returns "121005 ==> Mon Jan 23".

What I *think* (after some testing) is going on is that the Math::BigInt and Math::Bigfloat overload the arithmetic operators, which breaks timegm().

Is this expected behavior and/or documented anywhere?

In reply to Bignum breaks Time::Local? by cunningrat

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