Not true,

Since print $$reference (when ref $reference is SCALAR) will always output a value. When ref $reference is REF it will output SCALAR(0xXXXXXX).

The particular code I had in mind is:

  $a = "xyz";
  $b = \$a;
  $c = \$b;
  $d = $c;
  while (ref $d eq "REF") { $d = $$d; }
  $, = " & ";
  print ($a, $b, $c, $d, $dd);

[download]

If you look at the output then $d will have the same value of $d

What is different if there was no REF type, and you would need to have a reference (for whatever purpose) at the end of the while, then you will need to create a new reference which will not refer to the same scalar as $b:

  $a = "xyz";
  $b = \$a;
  $c = \$b;
  $d = $c;
  while (ref $d eq "SCALAR" or ref $d eq "REF") { $d = $$d; }
  $, = " & ";
  $e = \$d;
  print ($a, $b, $c, $d, $dd, $e);

[download]

$dd is ofcourse undefined, and $e refers to the scalar $d, not to $b. Which means you went one step to far in the iteration.

I don't know wheter this is useful or useless since I haven't really needed it yet, but if I need it I sure will be happy to have it. Else that second loop would need some extra code...

In reply to Re^3: ref == "REF" by Anonymous Monk
in thread ref == "REF" by gaal

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