variable and reference (later code samples I can not tell difference)

First comment:
Cloning
This is the process of replicating an object as it exists at the moment when the clone keyword is used in the script..  In other words, it creates a copy of an existing object $REFERENCE and assigns a new pointer $VARIABLE (variable name) to the copy.

// THIS CREATES A NEW POINTER (VARIABLE NAME) FOR THE SAME OBJECT

I thought a pointer was  a reference
But you wrote variable name.

Second comment:

 Each such entry in the symbol table is a reference to the box location.  

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I thought that
$x is variable
and
&$x is reference

but $x could reference box location?

Ray's comment:

Every variable name is a reference to a variable value (the information in the box) but not every use of the variable is done by reference.  For example, this procedural code creates two independent variables.  You can change $x and $y independently.

$x = 3;
$y = $x;

$x an $y are both variables which currently reference 2 boxes. Each box contains 3. But both can be changed independently.

This object-oriented code creates a variable (box with object inside) and two references to it.  You can use either name to change the object; the other name also refers to the changed variable.

$x = new StdClass;
$y = $x;

$x and $y are both variables which reference same box of new StdClass object.

Slick812 comment:
So a reference is not always a pointer but a variable points to a reference of a value?


until $c += 3;
$a,$b,$c are 10

<?php
$a = 5;
$b = &$a;
$b += 2;
///= = = = = = = = = = =
//or a ref to a ref -
$a = 5;
$b = &$a;
$b += 2;
$c = &$b;// $c is a reference to a reference
$c += 3;
//= = = = = = = = = = = 
class employee{
  public $employee;
  public function __construct($employee){
  $this->employee=$employee;
  }//end construct
}//end class
//IT IS MORE IMPORTANT, to learn WHY (for better code work) you would ever
//have NOT used the code line - 
$refEmp1 = new employee('Jay');
//BUT use this line - 
$refEmp1 = $emp1;  //Instead, what code work could be better if you used -
$refEmp1 = $emp1;   //instead? ?
//= = = = = = = = = = = = =
//and what code work would be better if you used -
$independentEmp1 = clone $emp1;
//instead of -
$independentEmp1 = new employee('Jay');

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Notice: Undefined variable: emp1 in C:\Users\Acer\Documents\NuSphere PhpED\Projects\noname325.php on line 23

Notice: Undefined variable: emp1 in C:\Users\Acer\Documents\NuSphere PhpED\Projects\noname325.php on line 24

Notice: Undefined variable: emp1 in C:\Users\Acer\Documents\NuSphere PhpED\Projects\noname325.php on line 27

Fatal error: __clone method called on non-object in C:\Users\Acer\Documents\NuSphere PhpED\Projects\noname325.php on line 27

As they say in My Fair Lady, "By Jove, I Think You've Got It!

Exactly right.

So a reference is not always a pointer but a variable points to a reference of a value?

I understand alittle better the difference

Thanks both.