Question

I was going to explain to our intern the difference between “pass by reference” and “pass by value” in PHP, and did this simple script:

$a=5; 
$b=&$a; 
$a=8; 
echo $b;
// prints  8
$a=5; 
$b=$a;  //no &
$a=8; 
echo $b;
// prints 5

However, running this in php-cli using php -qa yields:

php >     $a=5; 
php >     $b=&$a; 
php >     $a=8; 
php >     echo $b;
8
php >     // prints  8
php >     $a=5; 
php >     $b=$a;  //no &
php >     $a=8; 
php >     echo $b;
8
php >     // prints 5

Should not the $b=$a; unlink $a and $b?

… so I got curius, and tried:

php > $b=3;
php > echo $a;
3

So, how did I get this wrong? What’s going on here? It seems the reference-setting is somehow sticking, even though it should be cleared at the line $b=$a? I also tried:

php >    $e=5; $f=$e; $e=6; echo $f; 
5

…Which works as expected.

$a and $b seems linked permanently? Am I missing some big point here? How do I “unlink” the $a and $b variable?

Answer 1

Why should the reference be cleared if you assign a value to a variable? It works like this (with semi-simplified comments):

$a = 5;    // creates a "slot", puts 5 in it, makes $a point to it
$b =& $a;  // makes $b point to the same "slot" $a points to
$c = 6;    // creates a "slot", puts 6 in it, makes $c point to it
$a = $c;   // puts the value of the slot $c points to into the slot $a points to
echo $b;   // outputs the value of the slot $b points to (6)

It’s assigning a value to a variable. Whether the value is literal (5) or the value held by another variable doesn’t matter. The reference stays until you unset($b).