Reputation: 3391
A variation of this code passed by today (written by a perl coder), and it is confusing:
my $k = {3,5,6,8};
my $y = {%$k};
Why? What does that do? That seems to be the same thing as this:
my $y = $k;
The context is in a call using dbi module:
while (my $x = $db->fetchrow_hashref )
{ $y{something} = {%$x}; }
Upvotes: 4
Views: 343
Reputation:
The difference is that it's cloning the data structure without referencing the same memory.
For example:
use strict;
use warnings;
use Data::Dumper;
my $h={'a'=>1,'b'=>2};
my $exact_copy=$h; #$exact_copy references the same memory as $h
$h->{b}++; #$h maps b to 3
print Dumper($exact_copy) . "\n"; #a=>1,b=>3
my $clone={%$h}; #We dereference $h and then make a new reference
$h->{a}++; #h now maps a to 2
print Dumper($clone) . "\n"; #a=>1,b=>3 so this clone doesn't shadow $h
Incidentally, manually initialising a hash by using all commas (as in my $k = {3,5,6,8}
) is very, very ugly.
Upvotes: 8
Reputation: 385546
{ }
, in this case, is the hash constructor. It creates a new hash and returns a reference to it. So
Compare
my $k = { a => 4 };
my $y = $k;
$k->{a} = 5;
print $y->{a}; # 5
with
my $k = { a => 4 };
my $y = { %$k };
$k->{a} = 5;
print $y->{a}; # 4
Upvotes: 0