note 18376 deleted from ref.overload by sniper
| From: | sniper@php.net | Date: | Thu, 06 Feb 2003 14:19:57 +0000 |
| Subject: | note 18376 deleted from ref.overload by sniper | ||
| References: | 1 | Groups: | php.notes |
| Request: | Send a blank email to php-notes+get-43514@lists.php.net to get a copy of this message | ||
I think that object references should be implicit when using objects.
---
$my_object->child = $other_object;
$foo = $my_object->child;
$my_object->bar ( $other_object );
---
would be more interesting than
---
$my_object->child =& $other_object;
$foo =& $my_object->child;
class MyClass
{
...
function bar ( &$object ) {...}
...
};
---
__get, __set and __call will probably introduce a solution to this problem.
class MyClass
{
/**
* When accessing a class member, we return a reference
* to the member if it's an object. Else, we return a
* copy.
*/
function __get ($var_name, &$return)
{
if ( is_object($this->$var_name) )
{
$return =& $this->$var_name;
}
else
{
$return = $this->$var_name;
}
return true;
}
/**
* It seems that "__set()" does not take a reference to a
* variable as second parameter (cf: overload() example).
* But if &$value is possible, the following code should
* work.
*/
function __set ($var_name, &$value )
{
if ( is_object($value) )
{
$this->$var_name =& $value;
}
else
{
$this->$var_name = $value;
}
return true;
}
};
overload ("MyClass");
The __call() method should also give a way to get parameters' references if they are objects.
The problem :
$foo = $my_object->foo;
$foo is not a reference to $my_object->foo;
If the right member of the '=' sign is a reference, it would be great if '=' do
the same job as '=&'.