In bug #1269, Zeev says:

> There aren't too many exceptions to this case, basically, only
> three special functions may reference a variable without causing
> it to be implicitly declared - those are unset(), isset() and empty().

I seem to have found an instance where isset() will implicitly declare a variable.  I think.  Maybe it's the trinary operator doing it.  Either way, I don't think it should be happening.

  # php -q
  <?
  echo isset($x[0]);
  0
  $x[0] = isset($x[0]) ? 'y' : 'n';
  echo $x[0];
  y
  echo isset($x[0]);
  1

Obviously, $x[0] should end up being created when a value is assigned to it, but that should not happen until after the expression is evaluated to determine what value to assign.  A similar script that does not use arrays works as expected.  I spent an hour trying to figure out why my script wasn't working right when $x[0] was not set, since the code appears correct on inspection.  It just doesn't behave the same way it does with a simple scalar variable.

Dave