understanding issue with year got from localtime in perl - perl

Question

I have seen many blogs saying the localtime function returns years from 1900.
But for me when I add 180 days to the current date and print it using localtime it's showing the proper year as 2016 instead of 116.
Example:
$today_date=timelocal(0,0,0,27,03,2016);
$seconds=180*60*24*60;
$date=today_date+seconds;
$new_date=localtime(date);
print strftime("%d-%m-%Y",$new_date);
Output:
24-10-2016
When I give input as 116 instead of 2016 as year to timelocal It still gives the output as 2016.How is this possible?

Answer 1

Your sourcecode is wrong, you're missing the $ signs in front of the variable names.
You start by calling timelocal from the module Time::Local which takes the number of years since 1900 as input, but pass 2016 as year value. In the end, you pass this date back to localtime which is exactly the opposite function.
Try this:
print time;
print join(" ",localtime(time));
1461783061
1 51 20 27 3 116 3 117 1
You get:
Second 1
Minute 51
Hour 20
Day 27
Zero-based month 3 which is April
1900-based year 116 which is 2016
My default usage of localtime is:
my #now = localtime(time);
# Convert zero-based-month to calendar month
++$now[4];
# Convert 1900-based-year to calendar year
$now[5] += 1900;
If all you need is a date calculation, try Date::Calc
use Date::Calc qw(Today Add_Delta_Days);
print join '-', Today();
print join '-', Add_Delta_Days(Today(),1);
2016-4-27
2016-4-28
or DateTime:
use DateTime;
print DateTime->now->add(days => 1, seconds => 180);
2016-04-28T19:00:12
DateTime's strftime method or formatter option will give you any output you want.