CODEWITHSUNDEEP
arraysperlhashkey
boogie
boogie

Reputation: 66

What does $#vvv--; do to a hash in Perl?

This is the whole script, that for some mysterious to me reason outputs "642"

#!usr/bin/perl 
%vvv = (1,2,3,4,5,6);
$#vvv--;
%vvv = reverse %vvv;
print keys %vvv, "\n";

Also what does "keys" in the last statement do? Thanks for your time. I am just on hurry I don't have proper time to do my research. So again I appreciate you input.

Upvotes: 3

Views: 214

Answers (2)

TLP
TLP

Reputation: 67900

You should always run scripts with 

use strict;
use warnings;

If you had, you would have noticed an error:

Global symbol "@vvv" requires explicit package name at ...

Which means that $#vvv refers to the maximum index of the array @vvv, not the hash. In perl, @vvv and %vvv are two separate variables. So @vvv has nothing to do with %vvv, and that operation is without any use.

What the person who wrote the code might have been thinking of is a way to truncate an array:

my @array = 1 .. 6; # 1,2,3,4,5,6 but written as a range
$#array--;          # lower maximum index by 1
print "@array";     # prints "1 2 3 4 5"

However, that does not work wish hashes.

And as Friar has explained, reverse is a way to swap hash keys and values around. When used on a string, it reverses the string, e.g. "foobar" -> "raboof", but when used on a list, it reverses it, so 1,2,3,4,5,6 becomes 6,5,4,3,2,1.

Upvotes: 20

MadFriarAvelyn
MadFriarAvelyn

Reputation: 84

$#vvv-- looks like a comment. What's happening is the hash, being an even numbered element array, is just being reverse. So it goes from:

%vvv = (
    1 => 2,
    3 => 4,
    5 => 6
);

to:

%vvv = (
    6 => 5,
    4 => 3,
    2 => 1
);

So when the keys are printed, it grabs 642, or the new, current keys of the hash.

Upvotes: 6

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