On of the great things about YYYYMMDD is that if you sort it numerically you get it properly sorted. So just do

@ordered = sort { $a <=> $b } @dates;
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Comment: your expected output is as wrong as your output you get, they are identical... And in your code, $d is never defined. And the example is badly chosen as all parts are identical apart from the month. In this case, it does not matter, whether year is at the beginning or at the end.

UPDATE: Even if you sort string-wise you are fine, so sort @dates does the job.

One of the great things about YYYYMMDD is that if you sort it lexicographically you get it properly sorted!

@ordered = sort @dates;
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One of the great things about YYYYMMDD is that if you sort it lexicographically you get it properly sorted!

Yes! This is what bugs me about all the international date "standards" -- none of them sort naturally. The only tweak I use is for human readability, YYYYMMmmmDD, where mmm is the 3 letter abbreviation (in the language of choice), which helps the mildly dyslexic distinguish 5 March and 3 May (though maybe that's the hardest one).

-QM
--
Quantum Mechanics: The dreams stuff is made of

Please describe in your own words what the variables $c and $d are supposed to contain. Maybe print them out to verify your assumptions.

Have you noticed that dates in YYYYMMDD format can be sorted directly with sort? This is one of the major advantages of using YYYYMMDD as a date format.

Hi learner@perl,
Since your output 20130501 20130401 20130601 is the same with what you want what 20130501 20130401 20130601 what next? You have solved your own problems. Except your expected output is different.

Hi Masters.

I solved the problem

@dates = ('20130601', '20130401', '20130501');
my @sorted =
   map $_->[0],
   sort { $a->[1] cmp $b->[1] }
   map [ $_, join('', (/(..)(..)(....)/)[0,1,2]) ],
   @dates;  # DD-MM-YYYY
print "@sorted\n";
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Output

20130401 20130501 20130601
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Thanks all

Do you realize, that

join('', (/(..)(..)(....)/)[0,1,2])
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does not change a string $_ of length 8 in any way? (Apart from burning CPU cycles...). The regex also does not fit very well when splitting YYYYMMDD into pieces.

..I solved the problem..
Good... but why not just use sort just as you have been told previously on this post.
Yes your regexes matched successfully, see

Read more... (4 kB)
but even at that it a lot of work.
However, if you must use "Schwartzian transform", you can sort on the month, since the year is the same like this:

use warnings;
use strict;

my @dates = ('20130601', '20130401', '20130501');

print join ' ' => map{$_->[0]}
              sort{$a->[1] <=> $b->[1]}
              map{/.{4}(.{2})/;[$_,$1]} @dates;
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But sincerely, for this use sort just like other monks told you.
If you tell me, I'll forget.
If you show me, I'll remember.
if you involve me, I'll understand.
--- Author unknown to me

I've turned on warnings because I didn't get what's wrong with your code (which looks correct at first glance) and suddenly...

-bash$ xsel -o | perl -w  
Name "main::d" used only once: possible typo at - line 9.
Use of uninitialized value $d in numeric comparison (<=>) at - line 9.
Use of uninitialized value $d in numeric comparison (<=>) at - line 9.
20130601 20130501 20130401
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It looks like $d is never initialized in the first place, and evaluates to 0 in a numeric comparison. So to make life a bit easier, I would suggest always turning on strict and warnings:

#!/usr/bin/perl

use strict;
use warnings; # or just #!/usr/bin/perl -w on first line

my @dates = ('20130401', '20130501', '20130601');
my @ordered = sort { &compare } @dates;
print "@ordered\n";

sub compare {
    $a =~ /(\d{4})(\d{2})(\d{2})/;
    my $c = $3 . $2 . $1;
    $b =~ /(\d{4})(\d{2})(\d{2})/;
    my $d = $3 . $2 . $1;
    $c <=> $d;
}
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Don’t make this harder than it needs to be.   The reason for using a YYYYMMDD format is that it can “just be sorted as-is,” whether numerically or as a string.   Breaking it down into three groups, only to concatenate them in the same order, does nothing at all.   If you need to isolate the string from a bigger string, a simple pattern looking for 8 consecutive digits will do it.