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Posted by Chetan on July 25, 2008, 4:18 pm
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> Hi all
>
> As I understand it, object assignment causes the variable to contain
> the same object identifier, whereas assigning a variable via reference
> creates a true alias of the RHS:
>
> <?php
>
> class A
> {
> public $foo = 1;
> }
>
> $a = new A; //<id1>
> $b = $a; // <id1> = ($a) = ($b)
> $b->foo = 2;
> echo "2: ".$a->foo."\n"; // should print out 2
> echo "2: ".$b->foo."\n"; // should print out 2
>
> $c = new A; //<id2>
> $a = $c; // <id2> = ($a) = ($c), <id1> = ($b)
> echo "1: ".$a->foo."\n"; // should print out 1
> echo "2: ".$b->foo."\n"; // should print out 2
> echo "1: ".$c->foo."\n"; // should print out 1
>
> So far so good, as expected $a = $c causes $a to hold the same
> identifier as $c. If $a and $b were TRUE aliases, then echoing $b
> would print 1 (which it doesn't).
>
> Now, onto references
>
> Similar to the PHP manual, I group variables in parentheses that are
> true aliases of each other
>
> $A = new A; // <id3>
> $B = &$A; // ($A,$B) = <id3>
> $B->foo = 2;
> echo "2: ".$A->foo."\n"; // should print out 2
> echo "2: ".$B->foo."\n"; // should print out 2
>
> as expected
>
> echo "\n";
>
> $C = new A;
> $A = $C; // ($C) = ($A,$B) = <id3>
> echo "1: ".$A->foo."\n"; // should print out 1
> echo "1: ".$B->foo."\n"; // should print out 1
> echo "1: ".$C->foo."\n"; // should print out 1
>
> now, different to using assignment, changing $A to hold id3 means that
> $B will now hold id3, which means that all 3 should print out 1, which
> is what happens... still good.
>
> echo "\n";
>
> $D = new A; // <id4>
> $A = &$D; // ($A,$B,$D) = <id4>, ($C) = <id3> // shouldn't re-
> referncing $A also re-reference $B, as they are aliases of each other?
> // actually doing is ($B) = ($C) = <id3>, ($A,$D) = <id4>
> $A->foo = 2;
> echo "2: ".$A->foo."\n"; // should print out 2
> echo "2: ".$B->foo."\n"; // should print out 2
> echo "1: ".$C->foo."\n"; // should print out 1
> echo "2: ".$D->foo."\n"; // should print out 2
>
> What this actually prints out is 2 1 1 2. Now, when we write $A = &$D,
> as $A and $B are aliased, shouldn't they both then point to id4? What
> seems to be happening is that $B becomes un-aliased somehow...
>
> Taras
References in PHP can be confusing, depending on your notion of what they are
supposed to be. The best way to figure out seemed to be to look at the C source
code. It helps to know this is the way things are.
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