在PHP OOP中，对象分配没有意义

Question

I was tested and I got this wrong but this doesn't make sense:

class myClass
{
  public $x;

  function myMethod()
  {
    echo $this->x;
  }
}

$a = new myClass();
$a->x = 10;
$b = $a;
$b->x = 20;
$c = clone $b;
$c->x = 30;

$a->myMethod();
$b->myMethod();
$c->myMethod();

My intuition would be 102030 but the result is actually 202030!!! What happened to 10?!?! Shouldn't the variable of $a be left alone? I thought all objects are independent and would not be updated unless it has direct reference set by the ampersand (=&)?

Answer 1

In $b = $a;, only the object reference is being copied, not the object.

When you use clone, however, the object is indeed being copied, so $c = clone $b, creates both a new object (referenced by $c) and a new reference ($c).

In $b =& $a;, both symbols $a and $b would point to the same reference, that is, not even the reference would be copied (and therefore an assignment to $b of, say, an integer, would also affect the value of $a).

To sum up, there are two indirections here: from the symbol to the "zval" (in the case an object reference) and from the object reference to the object itself (i.e., to a portion of memory where the actual object state is stored).