Re: RE: PHP 4.0 Bug #6193 Updated: new object makes a shallow copy

You should think of "$a =& $b;" as "make the symbol table entry $a reference the same PHP value bucket as $b. If the $a symbol table entry was in use before it just changes it to point at the same PHP value bucket as $b. It is a symbol table re-assign. Look at the following example: function foo(&$a) {

        $b = 4;
        $a =& $b;

} foo($blah); In the beginning of the function $a is a local variable within foo() which points at the calling functions variable $blah. After the line "$a =& $b;" $a is re-assigned to point at $b, and therefore, disconnected from $blah. Andi At 11:10 PM 8/18/00 +0200, André Langhorst wrote:

PHP-DEV-POV:

They are not "double". They mean different things. This could be improved by adding some more logic into the compiler, and maybe will be. But it has nothing to do with compiler design.

from the mentioned POV they´re double

HD>>    - member(&x) { $x= &$this; } is broken.
HD>>            $x= &give_me_your_this(); // works
HD>>            member_function() { $x= &this; }; // fails

No it isn't a bug. Please read my previous mails.

I´d consider this a *very unexpected behaviour* (PHP-POV) $a=1; function foo(&$b) { $b=2; ... changes $a and $b to 2, why shouldn´t then $a=1; function foo(&$b) { $b=&$c; $a and $b and $c referenced afterwards? it really could need improvement regards --

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