This is a general problem with reference inside ternary operator. For ex., the following script fails with the same error:
$link = isset($i) ? (& $arr[$i]) : null;
- while the following works fine:
$link = & $arr[$i];

Thank you for taking the time to write to us, but this is not
a bug. Please double-check the documentation available at
http://www.php.net/manual/ and the instructions on how to report
a bug at http://bugs.php.net/how-to-report.php

Check the second note here:
http://php.net/manual/en/language.references.return.php

I thoroughly read the article you mentioned, Mike, but still don't understand why the following code fails to compile:

$link = isset($i) ? (& $arr[$i]) : null;
- while the following works fine:
$link = &$arr[$i];

In this case, &$arr[$i] is a legal reference assignment, so the first code should behave equal to
if (isset($i)) {
  $link = &$arr[$i];
} else {
  $link = null;
}
- but this code works fine, and mentioned above isn't even compiled. What's wrong with it?

I meant the documentation "Note:" (warning) not the user-contributed note.

Mike, I understand that. The second note tells I caanot return a reference to an expression result, such as &$object->method() or &(new StdClass()) - I can understand that. But the code sample I provided doesn't try to do that. To make things even simplier, the following code still fails to compile:
    $link = $flag ? &$a : &$b;
It doesn't try to return a reference to an expression, just a reference to a viriable; It doesn't try doing anything that the following code doesn't:
    if ($flag)
        $link = &$a;
    else
        $link = &$b;
And maybi I'm really stupid, but after 10 years in PHP development I still don't understand why the first code cannot be compiled :(

@marrch: PHP has no concept of a general & operator that takes the "reference" of a variable(whatever that's supposed to be). PHP only has a =& operator (which is really "=&" and not a combination of "=" and "&") which expects a variable on the right hand side. PHP also has a & modifier for function arguments and return values.

If you want to do yourself and others a favor, write $foo =& $bar rather than $foo = &$bar to make sure that there are no misunderstandings regarding this ;)

Thank you, Nikic! You've removed scales from my eyes. To my great pity and shame, I didn't understand that through all my working experience and really though & is a "take reference" operator, as one as exists in C/C++ etc :( Thank you once again for your explanation! Now I see this is not really a bug...