legLess has asked for the wisdom of the Perl Monks concerning the following question:

Monks ~

I'm trying (for readability's sake) to assign a branch of a nested hash to a new hash variable. I've solved my problem, but maybe I can decrease my Perl purity and solve it a different way next time. Here's a simplified version of the code:

# main nested hash
my %pages = (
    add_event_type => {
        show => {
            template => "add_evt",
            first_control => 'event_type_name',
            data => {
                event_type_name => { type => 'text', req => 1 },
                event_type_description => { type => 'text' },
            },
        },
    },
);

# elsewhere
my $page = "add_event_type";
my $action = "show";

# the code in question
my %page = %{ $pages{ $page }{ $action }};

# which allows, later
my $first_control = $page{ first_control };

# instead of 
my $first_control_again = $pages{ $page }{ $action }{ first_control };

[download]

--
man with no legs, inc.

Replies are listed 'Best First'.
Re: Getting a branch of a nested hash
by dragonchild (Archbishop) on Jan 27, 2003 at 18:48 UTC
    Looks good to me.

    Note: my %page = %{ ... }; is making a copy of the hash. Changes will not be retained. If you are changing that branch and want your changes to stick around, you will need to do something like my $page = ...; and $page->{first_control} = 'foo';

    As a rule of thumb, if you already have the reference, just use it, unless you don't want to change it and should copy it.

    ------
    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]
[select]
      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

[reply]
[d/l]
[select]
        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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