for(qw(4-10-31 14-11-31)) {
    /(\d+)(-.*)/ && printf "%d%s\n", 2000+$1, $2;
}
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I agree. In the "Real World", you probably would want to use a module like DateTime to do this sort of thing.

That one won't promote single-digit numbers to double-digit numbers.

s/(\d?\d)(-\d?\d)-(\d?\d)/(2000+$1).sprintf("%02g", $2).sprintf("%02g"
+,$3)/eg
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could do it, but that's similarly untested :-)

The listed examples didn't indicate to me that we need to worry about anything other than the first number, and adding 2000 to a 1 digit number definitely results in a 4 digit one. :-)

But if that's your spec, then try s/(\d?\d)-(\d?\d)-(\d?\d)/sprintf("20%02d-%02d-%02d", $1, $2, $3)/eg

for ( my @dates = qw( 4-10-31 14-11-31 ) ) {
        s/
        ^
        (\d)
        (\d)?
        (-\d{1,2}-\d{1,2})
        $
        /($2?"20$1":"200").$1.$+/ex
        and print;
}
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This formula is simpler than the one earlier. The main insight here is to avoid strings; it is better to use arithmetic addition to construct the 'year' string (i.e. 20xx) .

for ( my @dates = qw( 4-10-31 14-11-31 ) ) {
        s/
        ^
        (\d{1,2})
        ((?:-\d{1,2}){2})
        $
        /(2000+$1).$+/xe
        and print;
}
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I prefer to use strings - in that case, it becomes easy to deal with 3 or 4 digit years as well:

    my @dates = qw /4-10-31 14-11-31 1979-10-10/;
    for (@dates) {
        s/(\d+)/substr("2000", 0, -length($1)).$1/e;
        print "$_\n";
    }
    __END__
    2004-10-31
    2014-11-31
    1979-10-10
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Perl --((8:>*

The following will work correctly if the year is specified as 0, 1, 11, 111, or even 1111 (ones representing any digit):

use strict;
use warnings;

my( @dates ) = ( '4-10-31', '14-11-31' );

foreach my $date ( @dates ) {
    my $string = $date;
    $string =~ s/^(\d+)/
        my $year=2000;
        substr($year,4-length($1),length($1))=$1;
        $year/e;
    print $string, "\n";
}
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You can even override '2000' as being the base year by including the entire four-digit year, as in "1987-10-31".