my %h = (1, 2);
print bless(\%h), $/;
%h = undef;
print bless(\%h), $/;
__END__
main=HASH(0x99ae63c)
main=HASH(0x99ae63c)
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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;
}
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That way a new variable is generated for each iteration.

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 ?

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.

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.

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 ?

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.

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

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