How to make a shallow copy of a hash?

Question

I have a hash, which is filled up with data. I add it to an array. Then, I want to modify my hash, but make sure that what is in the array remains untouched. I'm trying this, doesn't work:

my @a;
my %h;
%h{foo} = 1;
push(@a, \%x);
%h = (); # here I clean the hash up
# but I want the array to still contain the data

I feel that I need to make a copy of it before adding to the array. How? I don't need a deep copy, my hash contains only strings.

The best solution is still to use a lexically scoped variable, and not try to reuse the same exact variable again. I'm sorry. — TLP, Dec 11 '22 at 17:32
You can make a shallow copy of a hash using assignment: `my %h2 = %h`, see also [this](https://stackoverflow.com/a/7083603/2173773) answer for copying a hash ref — Håkon Hægland, Dec 11 '22 at 18:00

Håkon Hægland · Accepted Answer · 2022-12-11T18:16:33.540

3

I don't need a deep copy, my hash contains only strings.

Then you could make a shallow copy of the hash:

my @a;
my %h;
%h{foo} = 1;
my %copy_of_h = %h;
push(@a, \%copy_of_h);
%h = ();

See also this answer for how to copy a hash ref. For more on Perl data structures, see perldsc and perldata.

edited Dec 11 '22 at 18:16

answered Dec 11 '22 at 18:09

Håkon Hægland

39,012
21
81
174

2

Or just `push @a, { %h }`. – TLP Dec 12 '22 at 00:01
1

@TLP Yes of course! No need to create a new lexical variable here – Håkon Hægland Dec 12 '22 at 09:15

How to make a shallow copy of a hash?

1 Answers1