abhishes has asked for the wisdom of the Perl Monks concerning the following question:

Hello Perl Monks,

I wrote the following program and the output really confused me. 

use warnings;
use strict;

my @a = ("a", "b", "c");
Foo(\@a);
print "$_ " foreach(@a);

sub Foo
{
  my $refA = shift;
  my @b = @$refA;
  $b[0] = "hello";
  $b[1] = "world";
  $b[2] = "Perl"; 
}

[download]

The output of the code was "a b c ".

How?? Didn't I pass a reference? If so, when I alter items against the reference, the change should be visible outside the subroutine as well!

I searched perl monks site found this node

but its still not clear to me... why were my changes lost.

regards,
Abhishek.

Replies are listed 'Best First'.
Re: Changing items of an array in a function
by Anonymous Monk on May 14, 2003 at 13:46 UTC

    Because you copied the referenced array's contents to a local copy and then modfied the local copy. You never copied it back. Try 

    use warnings;
use strict;

my @a = ("a", "b", "c");
Foo(\@a);
print "$_ " foreach(@a);

sub Foo
{
  my $refA = shift;
#  my @b = @$refA;
  $refA->[0] = "hello";
  $refA->[1] = "world";
  $refA->[2] = "Perl";
}

    [download]
[reply]
[d/l]
Re: Changing items of an array in a function
by broquaint (Abbot) on May 14, 2003 at 13:51 UTC
    How?? Didn't I pass a reference?
    That you did, but you also copied the contents of the array that reference pointed to into a new array, which as you can guess, isn't going to effect the array that said reference is pointing to. Here's what you really want 
    use warnings;
use strict;

my @a = ("a", "b", "c");
Foo(\@a);
print "$_ " foreach(@a);

sub Foo
{
  my $refA = shift;
  ## dereference the array
  $refA->[0] = "hello";
  $refA->[1] = "world";
  $refA->[2] = "Perl"; 
}

__output__

hello world Perl

    [download]
    It might be a good idea to learn more about references and how they work in perl. See. perlreftut, perlref, perldsc and tye's References quick reference for further information.
    HTH

    _________
    broquaint

[reply]
[d/l]
Re: Changing items of an array in a function
by edan (Curate) on May 14, 2003 at 13:52 UTC

    You did pass a reference, but then you created a new array: 

    my @b = @$refA;

    [download]

    And then you modify that new array which is only scoped inside the function, which leaves the array-by-reference argument as-is.

    If you change your function to look like this, it might do what you were expecting:

    sub Foo
{
    my $refA = shift;
    $refA->[0] = "hello";
    $refA->[1] = "world";
    $refA->[2] = "Perl";
}

    [download]
    --
    3dan
[reply]
[d/l]
[select]
Re: Changing items of an array in a function
by physi (Friar) on May 14, 2003 at 15:45 UTC
    You can also return your @b array to the main like that: 
    use warnings;
use strict;

my @a = ("a", "b", "c");
@a = Foo(\@a);
print "$_ " foreach(@a);

sub Foo
{
  my $refA = shift;
  my @b = @$refA;
  $b[0] = "hello";
  $b[1] = "world";
  $b[2] = "Perl";
  return @b;
}

    [download]
    -----------------------------------
--the good, the bad and the physi--
-----------------------------------
[reply]
[d/l]
[select]

Back to Seekers of Perl Wisdom