changing copy of a hash changes original

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

use strict;

sub fiddleHash($)
{
my ($ref_hash) = @_;
delete $ref_hash->{Quantity}{x};
}

sub displayHash($)
{
my ($ref_Hash) = @_;
foreach (keys %{$ref_Hash->{Quantity}})
  {
  print "$_:$ref_Hash->{Quantity}{$_}\n";
  }
}

my %hash1 = (
    Quantity => {
      x => 1,
      y => 2
      }
    );
my %hash2 = %hash1;

print "hash1 before call\n";
print displayHash(\%hash1);
print "hash2 before call\n";
print displayHash(\%hash2);
fiddleHash(\%hash2);
print "hash1 after call\n";
print displayHash(\%hash1);
print "hash2 after call\n";
print displayHash(\%hash2);
[download]

The sub fiddleHash is only supposed to alter hash2 but it also alters hash1
Output:

hash1 before call
y:2
x:1
hash2 before call
y:2
x:1
hash1 after call
y:2
hash2 after call
y:2
[download]

Comment on changing copy of a hash changes original Select or Download Code

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Re: changing copy of a hash changes original by poj (Abbot) on Nov 13, 2014 at 18:11 UTC
see Copy of multidimensional hash	[reply]
Re: changing copy of a hash changes original by CountZero (Bishop) on Nov 13, 2014 at 18:19 UTC
Add `print "$hash1{Quantity} - $hash2{Quantity}";` and you will see that they both have the same reference, in other words they are referencing the same anonymous hash. CountZero A program should be light and agile, its subroutines connected like a string of pearls. The spirit and intent of the program should be retained throughout. There should be neither too little or too much, neither needless loops nor useless variables, neither lack of structure nor overwhelming rigidity." - The Tao of Programming, 4.1 - Geoffrey James My blog: Imperial Deltronics	[reply] [d/l]
Re: changing copy of a hash changes original by 2teez (Vicar) on Nov 13, 2014 at 18:25 UTC
The sub fiddleHash is only supposed to alter hash2 but it also alters hash1 Since you use this `my %hash2 = %hash1;`, then do this `print %hash2; print %hash1;` [download] Then you will see why! If you tell me, I'll forget. If you show me, I'll remember. if you involve me, I'll understand. --- Author unknown to me	[reply] [d/l] [select]
Re^2: changing copy of a hash changes original by edo (Initiate) on Nov 13, 2014 at 18:44 UTC
Many thanks	[reply]
Re: changing copy of a hash changes original by Anonymous Monk on Nov 14, 2014 at 09:59 UTC
The problem is the hash copy because it's a shallow copy. %hash1 and %hash2 have a key by the name of 'Quantity' which happen to point to the same thing...namely {x=>1,x=>2}. In fact the fidleHash() sub doesn't change the hash which comes in as an argument, rather it changes the hash which the 'Quantity' key is pointing to. Insert the following two lines after copying the hash: my %hashCopy = %{$hash1{'Quantity'}} ; $hash2{'Quantity'} = \%hashCopy ; ...and you won't see the problem anymore. That way you also create a copy of what 'Quantity' points to.	[reply]
Re: changing copy of a hash changes original by locked_user sundialsvc4 (Abbot) on Nov 14, 2014 at 14:43 UTC
This is the difference between a so-called “shallow copy” and a “deep” one. Perl has the concept of references, which are scalar-things which “refer to” other things. (This is, by the way, how Perl appears to have “multi-dimensional,” even arbitrary, data-structures given that it actually only has scalars, lists, arrays, and hashes.) It is easy to “copy” something such that what you’re actually doing is creating a structure of references to the original thing ... easy, memory-efficient, and it usually works great. This is a “shallow” copy.