You could make a copy of the array in second(), but that wouldn't be very efficient if your array is large. If your real code is sufficiently similar to the example you could simply use a different array (or no array at all - just add values to the list that foreach iterates over). For example:

sub second {
    my $num_ref = shift;

    my $sum = 0;

    my @newarray = ( 6 .. 10 );

    # try one of these options
    # foreach ( @$num_ref, 6 .. 10 ) {
    foreach ( @$num_ref, @newarray ) {
    $sum += $_;
    }

    return $sum;
}
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HTH

Update: If you really do want to make a copy of the array, simply pass the array to the sub rather than passing an array ref:

my $result = second( @num );

sub second {
    my ( @array ) = @_;
# etc
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my $result = second( \@num );

sub second {
    my ( $aref ) = @_;
    my @new_array = @$aref;
    # my $new_ref = [ @$aref ]; # alternative
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Thanks, bobf

I need to pass the array as a reference because I'm passing more than one value. My example was simplified to make my question simpler.

It's a little bit harder to answer when your example is thus simplified. It looks like you got your answer, anyway.

(I might've suggested:)

sub second {
  my $sum = 0;
  $sum += $_ for 6, @{+shift};
  $sum
}
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I want to preserve the value of the original array @num after the operation at second. What is the best way to do that?

sub second {
   my $num_ref = shift;
    my @copy_of = @$num_ref;
   push(@copy_of, 6);

   my $sum = 0;

   foreach (@copy_of) {
      $sum += $_;
   }

   return $sum;
}
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Thanks, Darren :)

I need to copy the array into an array, and not make a copy of the reference, as in:

sub second {
   my $num_ref = shift;
   my $copy_of_ref = $num_ref;

   push(@$copy_of_ref, 6);

   my $sum = 0;

   foreach (@$copy_of_ref) {
      $sum += $_;
   }

   return $sum;
}
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Am I right?

Yes, that's fine. But please also take note of bobf's comments below :)

Update:Actually, I misread what you said there. Either way is okay. You can either take a copy of the reference, or dereference it into another array. But again, as bobf already pointed out in his update - the simplest way is to just pass the array (rather than a reference) - and then take a copy of that.

Hai, Don't dereference the original array. Instead of dereference just store it in different array in the second subroutine.

use strict;

my @num = qw(1 2 3 4 5);

first();

sub first {
   print "\@num before: @num\n";
   # prints 1 2 3 4 5
   my $result = second(\@num);
   print "result: $result\n";
   print "\@num after: @num\n";
   # prints 1 2 3 4 5 6
   # But I would need it to still print
   # 1 2 3 4 5
}

sub second {
   my @num_ref = shift;

   push(@num_ref, 6);

   my $sum = 0;

   foreach (@num_ref) {
      $sum += $_;
   }

   return $sum;
}
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