in reply to Re: Getting a branch of a nested hash
in thread Getting a branch of a nested hash

Note: my %page = %{ ... }; is making a copy of the hash. Changes will not be retained.
Are you sure? Try running this: 
#!/usr/bin/perl

use strict;
use warnings;
use Data::Dumper;

my %hash = (key1 => {
                     subkey1 => [ qw(a number of elements in an array)
+ ],
                    },
            key2 => "value2");

my %hash2 = %hash;

$hash2{key1}->{subkey1} = [ qw(other elements in an array) ];
print Dumper(\%hash);
print Dumper(\%hash2);

[download]
A hash copy will only do a shallow copy, i.e. the references are copied instead of the contents of those references. That's why changes to %hash2 also end up in %hash. Using Devel::Peek to display %hash and %hash2 will make this even more clear. Search for the lines looking like: 

  Elt "key2" HASH = 0x3e4d4a
    SV = PV(0x80f6498) at 0x81009c4

[download]

and note the addresses.

Arjen

Replies are listed 'Best First'.
Re3: Getting a branch of a nested hash
by dragonchild (Archbishop) on Jan 27, 2003 at 22:24 UTC
    Kinda sorta not really. It all depends on whether what you're modifying is the stuff in the hash you copied or in child hashes. Let's say you have: 
    my %hash = (
    key1 => {
        a => 1,
        b => 2,
    },
    key2 => 'blah'
);

my %hash1 = %{ $hash{key1} };

$hash1{a} = 3;    # This change persists
$hash1{c} = 5;    # This change does NOT persist

    [download]
    It all depends, doesn't it? :-) (And also why objects should encapsulate this crap.)

    ------
    We are the carpenters and bricklayers of the Information Age.

    Don't go borrowing trouble. For programmers, this means Worry only about what you need to implement.

[reply]
[d/l]

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