use strict;
use warnings;
use Time::Piece;

my $month = 11;
my $t = Time::Piece->strptime($month, '%m');
my $last_day_month = $t->month_last_day();
print "$last_day_month\n";

__END__

30
[download]

To account for leap years, you may need to also specify the year.

my $last_day_month = $t->month_last_day($month); #??
[download]

But you can chain the method calls:

$last_day = Time::Piece->new->month_last_day;
[download]

What do you mean when you say # Create Time::Piece New Object for the dates? It would be helpful to know what you are trying to accomplish overall. The new() method without arguments creates an object for "now", and if you pass other attributes in but not the year, it will assume you mean the current year. So if you loop through the months and create a T::P object for each one, you will get the last day of the month for months in the current year. This is not accurate if you use the list for other years: as toolic said, due to leap years you will need pass the year to the object creator if you want your code to correctly handle years other than the current year.

#!/usr/bin/perl
use strict;
use warnings;
use feature qw/ say /;
use Time::Piece;

for my $year (2015, 2016) {
  say sprintf("$year/%02d: ", $_), Time::Piece->strptime("$year $_", "
+%Y %m")->month_last_day) for 1,2;
}

__END__
[download]

$ perl 1140366.pl
2015/01: 31
2015/02: 28
2016/01: 31
2016/02: 29
$
[download]

A bit obscure

$t->add_months(11-$t->mon)->month_last_day;
[download]

clearer

use Date::Calc qw( Days_in_Month );
print Days_in_Month(2016,2);
[download]