Case three and four is wrong. Naming changing nothing about this. What you can do, is register _specific_ references as having a name, _and_ name the root objects. However this is not feasable unless you know what these references are ahead of time. For a different and IMO clearer example of this problem try the following: 

Dumper version: 2.102
----------------------------------------
Before:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

After:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

$dump and $after are the same as each other

Before:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

After:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

$dump and $after are the same as each other

Before:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

After:
$VAR1 = [
          \\do{my $o},
          do{my $o}
        ];
${${$VAR1->[0]}} = $VAR1->[0];
$VAR1->[1] = ${$VAR1->[0]};

$dump and $after are the same as each other

----------------------------------------
$VAR1 = [
          \'Test',
          'Test'
        ];

$VAR1 = [
          \'Test',
          ${$VAR1->[0]}
        ];

$VAR1 = [
          \'Test',
          ${$VAR1->[0]}
        ];

[download]

---
demerphq

    First they ignore you, then they laugh at you, then they fight you, then you win.
    -- Gandhi

• Update:  
*blush* the before and after was wrong. Still my point is clear. These data structures are different, but dump the same.


In reply to Re: Re: Data::Dumper scalar refs undef by demerphq
in thread Data::Dumper scalar refs undef by bsb

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