I've been reading about memory management in PHP and learned that variables in PHP copy the reference to zvals as long as you don't do a write operation (copy on write paradigm).
However, it does not describe what happens when you reassign a value to an already copied zval.
Here's what the book says:
$a = 1; // $a = zval_1(value=1, refcount=1)
$b = $a; // $a = $b = zval_1(value=1, refcount=2)
$c = $b; // $a = $b = $c = zval_1(value=1, refcount=3)
$a++; // $b = $c = zval_1(value=1, refcount=2)
// $a = zval_2(value=2, refcount=1)
Now, if you do another $a++, will it change the value in zval,
$a++; // $a = zval_2(value=3, refcount=1)
or will it create a new zval once again?
$a++; // $a = zval_3(value=3, refcount=1)
Following the logic of the PHP Language Reference, I guess it should be more likely option #1 as long as refcount = 1 (you would not manipulate another variable).
