PHP Object Assignment vs Cloning

Question

I know this is covered in the php docs but I got confused with this issue .

From the php docs :

$instance = new SimpleClass(); $assigned   =  $instance; $reference  =& $instance; $instance->var = '$assigned will have this value'; $instance = null; // $instance and $reference become null var_dump($instance); var_dump($reference); var_dump($assigned); ?> 

The above example will output:

NULL NULL object(SimpleClass)#1 (1) { ["var"]=>  string(30) "$assigned will have this value" } 

OK so I see that $assigned survived the original object ($instance) being assigned to null, so obviously $assigned isn't a reference but a copy of $instance.

So what is the difference between

 $assigned = $instance  

and

 $assigned = clone $instance 

Answer 1

Objects are abstract data in memory. A variable always holds a reference to this data in memory. Imagine that $foo = new Bar creates an object instance of Bar somewhere in memory, assigns it some id #42, and $foo now holds this #42 as reference to this object. Assigning this reference to other variables by reference or normally works the same as with any other values. Many variables can hold a copy of this reference, but all point to the same object.

clone explicitly creates a copy of the object itself, not just of the reference that points to the object.

$foo = new Bar;   // $foo holds a reference to an instance of Bar $bar = $foo;      // $bar holds a copy of the reference to the instance of Bar $baz =& $foo;     // $baz references the same reference to the instance of Bar as $foo 

Just don't confuse "reference" as in =& with "reference" as in object identifier.

$blarg = clone $foo;  // the instance of Bar that $foo referenced was copied                       // into a new instance of Bar and $blarg now holds a reference                       // to that new instance