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.

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.