Re: anon ref to scalar?

[drieux]

On Dec 25, 2003, at 5:02 AM, [EMAIL PROTECTED] wrote:

Here's the code which I expected to work, but doesn't.
I'm expecting \$$orig to result in a new reference, but
instead it's reusing the original reference that I started with.
So:

    $s = "some string";
    $orig = \$s;
    $new = \$$orig;
    print "orig:$orig, new:$new\n";
Results in these values:

    orig:SCALAR(0x17753dc), new:SCALAR(0x17753dc)
bear with me, I am slow this morning.

if you went with "$new = $$orig" then your new
holds the value that $s holds. Hence it would
not be a 'reference' to that string...
Plan A: would seem to be the simple

        my $bob = $$orig; #copy the substance of $orig
        my $new = \$bob;  #take a reference to the new memory
Plan B: would be to make a function like:

        #------------------------
        #
        sub clone_me
        {
                my ($me,$item) = @_;
                
                my $ref_type = ref($item);
                
                return $item unless $ref_type;
                
                if ( $ref_type eq 'SCALAR' )
                {
                        my $new_item = $$item;
                        return(\$new_item);
                }elsif ($ref_type eq 'ARRAY' )
                {
                        my $count = 0;
                        my @array;
                        $array[$count++] = $me->clone_me($_)
                                foreach(@$item);
                        
                        return([EMAIL PROTECTED]);
                }
                # solve the case for a hash ref
                                
        } # end of clone_me
ciao
drieux
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