Passing by reference, $a = &$b, is $a=$n the same as $b=$n?

Question

I'm trying to understand passing by reference. This is probably a bad analogy, but is it like Newton's 3rd law (action-reaction pairs)?

For example, for the following code

$a = 4;
$b = 2;
$n = 42;
$a = &$b;

is

$a=$n the same as $b=$n? Isn't the value of $a and $b stored in the same address?

Answer 1

If you assign these variables normally:

$a = 1;
$b = 2;
$c = 3;

Then the associations will look like this:

a --> 1
b --> 2
c --> 3

Now, if you make $c into a reference of $a, like this:

$a = 1;
$b = 2;
$c = &$a;

Then the associations will look like this:

a --> 1 <--.
b --> 2    |
c --------/

In other words, $a and $c point to the same value. Because they both point to the same value we can change either of the variables and they will both point to the new value.

$a = 5;
echo "$a $c"; // Output: "5 5"
$c = 10;
echo "$a $c"; // Output: "10 10"

Answer 2

Yes, After assigning $n to $a, $b will point $n.

This is because after executing $a=&$b both $a and $b references a same memory location and become reference variable (is_ref=1). And the reference count (refcount) of that specific memory location increases by 1. Now whatever value you assign to any of those references both will point to same value.

Executing $a=$n means value of $n will be store to the location referenced by $a. And this is same location as $b.

See the example here.

$a, $b, $n are pointing different locations

php > $a = 4;
php > $b = 2;
php > xdebug_debug_zval('a'); // they are pointing different location
a: (refcount=1, is_ref=0)=int(4)

php > xdebug_debug_zval('b'); // they are pointing different location
b: (refcount=1, is_ref=0)=int(2)

php > $n = 42;
php > xdebug_debug_zval('n'); 
n: (refcount=1, is_ref=0)=int(42)

$a and $b both becomes a reference now

php > $a = &$b; 
php > xdebug_debug_zval('b');
b: (refcount=2, is_ref=1)=int(2)

php > xdebug_debug_zval('a'); // a too
a: (refcount=2, is_ref=1)=int(2)

Assigning new value, NOT references to any of $a and $b

php > $a = $n;
php > xdebug_debug_zval('a'); // a holds $n's value '42' now
a: (refcount=2, is_ref=1)=int(42)


php > xdebug_debug_zval('b'); // same for b
b: (refcount=2, is_ref=1)=int(42)