Foreach behavior with variables by reference

Question

I was doing some tests and I realized that the foreach has a strange behavior.

Suppose I have the following array:

$array = array('primeiro', 'segundo', 'terceiro');

When I run one foreach using a reference, and then another without reference foreach (with the same variable), see what happens:

$array = array('primeiro', 'segundo', 'terceiro');

foreach ($array as &$b);

var_dump($array);

foreach ($array as $b);

var_dump($array);

Upshot:

array (size=3)
  0 => string 'primeiro' (length=8)
  1 => string 'segundo' (length=7)
  2 => &string 'terceiro' (length=8)

array (size=3)
  0 => string 'primeiro' (length=8)
  1 => string 'segundo' (length=7)
  2 => &string 'segundo' (length=7)

See that the last element has a different value.

Then two questions arose:

• The last value of $array has become the penultimate element?
• Why, after a foreach with variable by reference, the last element is as referenced by the variable (which is after the as)?

Observing: Repair the sign of & in the result.

Answer 1

When you "equal" two variables, you can do this in two ways:

• By value: Indicating that at that point in the code the value of one is equal to that of the other:

  $a = $b;
  

• By reference: Indicating that, until it is said otherwise, they are equal, that is, what happens to one should happen to the other:

  $b = "b";
  $a = &$b;
  var_dump($a);   //Imprime: var 'b'
  
  $a = 'a';
  var_dump($b);   //Imprime: var 'a'
  

When passing the variable $b by reference in the first foreach, you are, at each iteration, linking the value of $b with a vector position $array.

$b = &$array[0];  //na primeira iterada 

When establishing the link with another variable, the first link is broken, that is, in the second iteration $b is linked with $arrray[1] and no longer with $array[0]. In other words, at the end of the first foreach, $b is linked to (and only to) the third position of the vector:

$b = &$array[2];  //na última iterada

In the second foreach, with each iteration, we are saying that $b has a new value and therefore, $arrray[2] also has a new value with each iteration. In the penultimate iteration we do:

$b = $array[1];
//o que implica que também estamos fazendo $array[2] = $array[1];

And therefore, when we arrive at the last iteration of the second foreach, the last position is with the value of the penultimate.