Perl uses reference counting. Every time the code goes over a "my" variable, you are going to get new memory unless Perl can reuse what it had before (and it will it if it can - but Perl doesn't do garbage collection like JAVA).

If you pass a reference to that memory from the: my $trick = "TRICKY"; statement out of a sub, the next time Perl sees the "my", it allocates new memory for it if it sees that what it did before is still in "use" - meaning a reference to it exists. Perl "frees" memory back for its own use when it sees that the reference count to that memory is zero - not well illustrated in this example. The point here is that you get a completely new copy of "TRICKY" every time the sub is called.

My verbage was a bit confusing, but does the code answer the question?

#!/usr/bin/perl
use strict;
use warnings;

my @tricks;

for (1..3)
{
   my $ref_to_a_trick = get_a_trick();
   push @tricks, $ref_to_a_trick;
}

print @tricks,"\n"; #SCALAR(0x182b394)SCALAR(0x24920c)SCALAR(0x24925c)
#note each "trick reference" points to a different location

#@tricks is an array of scalar references

foreach my $ref (@tricks)
{
    print "$$ref\n";  #de-reference each "trick" to get the value
    #TRICKY
    #TRICKY
    #TRICKY
}

sub get_a_trick
{
   my $trick = "TRICKY";
   return (\$trick);
}
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@tricks=(); will "free" the $tricks memory back to Perl (not to the O/S) and Perl will reuse it if it can because no references exists any longer to various memory allocations of "TRICKY".