That's a choice quote. Put another way, the OP is suffering from some confusion that if (defined $x) is intended to be the same as if ($x). In fact, they serve different purposes.

defined tests whether a variable has any value -- a false value is still a value.

Short demonstration code: 

my $x;
demonstrate();

$x = 1;
demonstrate();

$x = 0;
demonstrate();

sub demonstrate {
  if (defined $x) { print "\$x has value $x\n"; }
  else { print "\$x is undefined\n"; }

  if ($x) { print "\$x evaluates as TRUE\n"; }
  else { print "\$x evaluates as FALSE\n"; }

  print "\n";
}

[download]

This produces the following output: 

$x is undefined
$x evaluates as FALSE

$x has value 1
$x evaluates as TRUE

$x has value 0
$x evaluates as FALSE

[download]
<–radiant.matrix–>
Ramblings and references
The Code that can be seen is not the true Code
I haven't found a problem yet that can't be solved by a well-placed trebuchet

In reply to Re^2: if ($2) behaves differently than if (defined $2) by radiantmatrix
in thread if ($2) behaves differently than if (defined $2) by walto

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