hash refferences

[leonidlm]

Hi all
Just wonder, take a look at the following sample code:

my @arrayOfHashes;
my %temporary;

foreach $val (@variables) {
# Here I populate a temporary hash
#
# ....
#

push @arrayOfHashes, \%temporary;
%temporary = undef;
}

My question is why after I assign a reference to an object and then I loose the original reference to it (by assigning another value -> undef) The original hash in the array is changed too!
Thanks.

[Corion]

The reference still points to the same place as %temporary, so changing one will also change the other. I suggest the following code:

foreach my $val (@variables) {
    my %temporary = ...;

    push @arrayOfHashes, \%temporary;
};

%temporary is not a reference and hence you can't "lose" the reference to its data.

[moritz]

Assigning undef to a hash doesn't "destroy" the hash. I t just empties it.

my %h = (1, 2);
print bless(\%h), $/;
%h = undef;
print bless(\%h), $/;
__END__
main=HASH(0x99ae63c)
main=HASH(0x99ae63c)

It's the same memory address after assigning undef.

The usual way to do it is to declare the hash inside the loop:

for $val (@variables) {
   my %temp;
   # work with %temp here;

   push @array, \%temp;
}

That way a new variable is generated for each iteration.

[leonidlm]

But whats the difference between the following:
1. Declaration of the %temp inside the loop
2. Performing: undef %temp at the end of each iteration
Someone ?

[Corion]

They are completely different. undef %temp empties the variable but still keeps the variable there, while a lexical variable goes out of scope. The undef %temp acts on the values, while leaving the scope acts on the binding of the name to the value.

[moritz]

What Corion said is right.

%temp is actually a container - think of a bucket. You can fill this container (putting water into the bucket), and empty it, which is what %temp = undef does.

Or you can but your bucket into the cupboard, (pushing it to an array), and take a new bucket (declaring %temp inside the loop).

In this real-world analogy you can see that there's a huge difference between these two things.

[leonidlm]

Another thing: Unfortunately I can't release the %temp variable every loop iteration, I need to perform it on demand. How I can do it ?

[moritz]

push @array, { %temp };

That actually copies the items from %temp into an anonymous hash, and returns its reference.

See perlreftut and perlref for more details.

[Corion]

You can't. You can store copies of the values in another variable maybe.