The post from haukex is excellent++.

When I looked at your "internet code", it does functionally exactly the same thing every time it runs.
It will indeed allocate and use a new hunk of memory and return a reference to that new memory for each call, but the results in the created @array are exactly the same.

I wrote some more demo code for you below.

If return_array() is to produce different results upon each call, it could be that calculate() is intended to cause that result? If so, then the code as shown won't do that. Besides understanding the basics of references, the aliasing nature of a Perl for loop is also important to understand. I show a couple of ways to iterate over each element of @array, modifying @array in the process, before returning the reference to @array to the caller.

I hope that this is instructive on what is actually a pretty complicated subject. 

#!/usr/bin/perl
use strict;
use warnings;
use Data::Dump qw(pp);
$| =1;  #some advanced stuff to make the printout in the right order
        #forget about this for now...

# a reference to an array is a scalar value..

my $ref2_array_first_run  = return_array(1);
my $ref2_array_second_run = return_array(4);

print "Scalar value of \$ref2_array_first_run is $ref2_array_first_run
+\n";
pp $ref2_array_first_run;

print "Scalar value of \$ref2_array_second_run is $ref2_array_second_r
+un\n";
pp $ref2_array_second_run;

my @copy = @$ref2_array_first_run;  # copy array to another block of
                                    # memory called @copy
$ref2_array_first_run = undef;      # zap memory that first call to
                                    # return_array() allocated.
pp $ref2_array_first_run;           # memory no long in use!

pp \@copy;                          # A copy of the first run now
                                    # exists in other memory


sub return_array 
{
    my $start_num = shift; #so that each run is different
    
    # my @array = (1, 2, 3);
    my @array = ($start_num..$start_num+2);
    
    # Question is whether or not calculate() should modify @array?
    # If so, here are 2 ways to do that...
    
    foreach my $element (@array) {
    
        # calculate($element); # does not modify @array!
        # these 2 calculate methods both modify @array...
        
        $element = calculate($element); # most obvious way
        calculate2 (\$element);         # using ref to $element
    }
    return \@array;
}

sub calculate
{
   my $value = shift;
   return ($value *5);  #multiplies by 5 
}

sub calculate2
{
   my $ref2scalar = shift;
   $$ref2scalar *= 10;  # multiplies by 10 
                        # no "return value" needed
}
__END__
Prints:
## This shows that the ARRAY references created by run1 and run2
## do indeed point to different memory

Scalar value of $ref2_array_first_run is ARRAY(0x6c35e0)
[50, 100, 150]  # "pretty print" output for 1,2,3
Scalar value of $ref2_array_second_run is ARRAY(0xfcb430)
[200, 250, 300] #  "pretty_print" output for 4,5,6
undef           # memory used for the first array creation is GONE.
[50, 100, 150]  # the @copy of the first run still exists

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In reply to Re: returning reference of a variable defined inside a subroutine. by Marshall
in thread returning reference of a variable defined inside a subroutine. by aswingeo

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