ID: 47967
Comment by: code at dumb dot ro
Reported By: r at dumb dot ro
Status: Open
Bug Type: *General Issues
Operating System: all
PHP Version: 5.2.9
New Comment:
One more thing I'd like to add:
In order for this function to work properly and for cloning in general,
I think PHP would need 2 different kinds of references, an implicit and
an explicit one.
Basically if I'd say $arr = array(new Obj()); it would use an implicit
reference, therefore when I'll clone the array the object would also be
cloned. If I say $arr = array(&new Obj()); that would be an explicit
reference so I want the array to be an array of references and therefore
the referenced object won't be cloned.
This subject should be discussed in detail before implementation,
because it has huge ramifications, basically the whole cloning process
could be changed.
Previous Comments:
------------------------------------------------------------------------
[2009-04-14 14:44:49] r at dumb dot ro
Description:
------------
When you pass an array of objects to a function (not by reference) and
you modfiy one of the objects in the array, the main array (the
referenced objects) will be modified. This wasn't a problem until
objects got passed by reference by default.
Reproduce code:
---------------
$class A
{
}
$a = new A();
$a->x = 10;
$arr = array($a);
function foo($arr)
{
$arr[0]->x = 1;
}
foo($arr);
print_r($arr);
Expected result:
----------------
the expected result would be 10
Actual result:
--------------
since the array is internally using references for the stored objects
the result will be 1.
to fix this I wrote a function that clones the array properly and I'm
calling the function like this:
foo(array_clone($arr));
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Edit this bug report at http://bugs.php.net/?id=47967&edit=1
