Re: the last added element of a hash

[Athanasius]

Hello Lennotoecom,

I would re-write this idea as follows:

#! perl
use strict;
use warnings;
use List::Util 'max';

my %h =
(
   '11c'  => 'C',
    b     => 'B',
    e22   => 'E',
    aaaaa => 'AAAA',
);

if (%h)
{
    my $last = max map { \$_ } values %h;
    printf "The last element added to the hash was %s\n", $$last;
}
else
{
    print "No elements have been added to the hash\n";
}

This gives the same result, but:

• compiles under use strict;
• runs without warnings under use warnings; and
• handles the case where the hash is empty.

This script’s output (like the title of this node) is still somewhat misleading, however, as “last element added” really means “last element added of those still remaining in the hash.” For example, if you add elements X, Y, and Z, and then delete element Z, this script will output Y as the “last element added,” which isn’t strictly correct.

But the real problem with this whole strategy is that it relies on an implementation detail of the perl interpreter; namely, that it always assigns new hash values to memory addresses above those of previously-assigned values. Is this guaranteed to be the case? Will it always be true of perl implementations in the future? I don’t know, and for that reason I would advise against this approach until and unless it can be proven to be valid now and also future-proof.

Hope that helps,

[LanX]

> always assigns new hash values to memory addresses above those of previously-assigned values. Is this guaranteed to be the case? 

No, as soon as an address is freed again it might be reused.

Just try to delete an entry and create new ones ...:)

update

Not a meditation and poorly worded!

moved to SOPW.

[Discipulus]

In fact this smell like things you never have to attempt..
good point LanX!

But you Lennotoecom, what do you think about? your meditation is so poor in words...

Can i add: when you play with memory's addresses in Perl you are always going in a wrong direction.

L*