hash dereferencing / copying

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

Patient ones:

Perl surprised me again this morning. I wanted to *copy* an anonymous hash by dereferencing it, and then creating a new reference to it, leaving me with two references to two different hashes. So I did:

my $this = { 'name' => 'alex' };
my $that = \%{ $this };
# now i thought that $that would be an independent copy, but:
$that->{'name'} = 'madonna';
print $this->{'name'}; 
# oh no! i'm madonna! they both referenced the same thing
[download]

This surprised me, because I thought dereferencing would make it just a 'free-floating' anonymous hash. This works fine, but is an extra line:

my $this = { 'name' => 'alex' };
my %that = %{ $this };
my $that = \%that;
# now $that really is an independent copy, and:
$that->{'name'} = 'madonna';
print $this->{'name'}; #alex
# phew! I'm still me!
[download]

My questions are: 1) why does this happen like this? I don't really see what the difference is between the first and the second code sample. What subtlety am I missing? 2) Is there a neater way of writing the second sample, which does what I want, without using the throwaway named hash %that?

with thanks

/=\

Comment on hash dereferencing / copying Select or Download Code

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Re: hash dereferencing / copying by broquaint (Abbot) on Feb 27, 2002 at 12:59 UTC
In your first example you're essentially copying the reference. You probably want to do something like this `$this = { foo => bar }; $that = { %$this }; $that->{foo} = "not bar"; print $this->{foo}; __END__ bar` [download] This works because it creates a new anonymous hash and assigns it to `$this`, whereas in your first example you're just dereferencing `$this` and then assigning a reference to it (still with me ;-). So in your second example you're creating a new hash in `%that` and then assigning a reference to it, which is why it doesn't effect `$this`. HTH broquaint	[reply] [d/l]
Re: hash dereferencing / copying by strat (Canon) on Feb 27, 2002 at 13:25 UTC
For more complex datastructures, I like to use the perl-Module Storable to copy the content... Best regards, perl -le "s==F=e=>y~\martinF~stronat~=>s~[^\w]~~g=>chop,print"	[reply]
Re: Re: hash dereferencing / copying by PrakashK (Pilgrim) on Feb 27, 2002 at 15:34 UTC
Or, Data::Dumper (it is slower than Storable though). `use Data::Dumper; my $this = {a => 1, b => {bb => [ 'ccc' ]}, c => 3}; # clone $this into $that my $that = eval Data::Dumper::Dumper($a);` [download] /prakash	[reply] [d/l]
Re: Re: Re: hash dereferencing / copying by IlyaM (Parson) on Feb 27, 2002 at 23:18 UTC
Or use Clone. It is faster than Storable and Data::Dumper and more reliable than Data::Dumper. -- Ilya Martynov (http://martynov.org/)	[reply]
Re: hash dereferencing / copying by Ido (Hermit) on Feb 27, 2002 at 13:36 UTC
The code `my $that={%$this}` will do for copying one level of references. But if you want more than one level, it won't: `my $this={name=>'alex',array=>['one','two']}; my $that={%$this}; $that->{name}='Ido'; print $this->{name}; #alex - not affected $that->{array}[0]='three'; print $this->{array}[0];#three - affected` [download] In order to copy more than one level of references, use Data::Dumper: `use Data::Dumper; my $this={name=>'alex',array=>['one','two']}; my $that=eval Dumper $this; $that->{name}='Ido'; print $this->{name}; #alex - not affected $that->{array}[0]='three'; print $this->{array}[0];#one - not affected` [download]	[reply] [d/l] [select]