The following snippet of perl code behaves in a manner that I do not understand. I realize there is another approach but I want to learn what perl is doing in this snippet (The entire test program is at the end of the post) : 
foreach $ralias (@{$refarrayptr})
{
    $raliashash = $ralias;
    
    $reference = $ref{$raliashash};

    $ralias = ${$reference}[0];
}

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$refarrayptr is a reference to an array containing hash keys. $ref is a reference to a hash containing references to anonymous arrays. Using the x command in the perl debugger, $refarrayptr looks as follows: 
  DB<43> x $refarrayptr
0  ARRAY(0x9bfef4c)
   0  'key'
  DB<44>

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Once $ralias has been reassigned with the value of the first element of the $reference array, $ralias doesn't have the value of the first element and the first element of $refarrayptr changes from 'key' to the first element of the $reference array as seen below once again in the debug output: 
  
DB<48> x $refarrayptr
0  ARRAY(0x9bfef4c)
   0  'a'
DB<49>

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Here is the output of the code: 
./testprob.pl
Initial content of ARRAY(0x9ead350) -> key
Final content of ARRAY(0x9ead350) -> a

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$ralias is a simple scalar (I think). Why is the assignment to $ralias changing the $refarrayptr array since it is a simple scalar. Thank you 
#!/usr/bin/perl

@list = qw(a b c d);

$reflist = \@list;

#$ref{$reflist} = $reflist;
$ref{'key'} = $reflist;

push(@refarray,keys %ref);

$refarrayptr = \@refarray;

foreach $ralias (@{$refarrayptr})
{
    print "Initial content of $refarrayptr -> " . @{$refarrayptr}[0] .
+ "\n";
    $raliashash = $ralias;
    
    $reference = $ref{$raliashash};

    $ralias = ${$reference}[0];
    
    print "Final content of $refarrayptr -> " . @{$refarrayptr}[0] . "
+\n";
}

[download]

In reply to assignment to scalar changing array content by colemannist

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