PHP Pass by reference error after using same var

Question

Take a look to this code, and help me to understand the result

$x = array(\'hello\', \'beautiful\', \'world\');
$y = array(\'bye bye\',\'world\', \'harsh\');         


        

Answer 1

// ...
$v = "DONT CHANGE!";
unset($v);
// ...

because $v is still a reference, which later takes the last item in the last foreach loop.

EDIT: See the reference where it reads (in a code block)

unset($value); // break the reference with the last element

Answer 2

Foreach loops are not functions.An ampersand(&) at foreach does not work to preserve the values like at functions. So even if you have $var in the second foreach () do not expect it to be like a "ghost" out of the loop.

Answer 3

After this loop is executed:

foreach ($x as $n => &$v) { }

$v ends up as a reference to $x[2]. Whatever you assign to $v actually gets assigned $x[2]. So at each iteration of the second loop:

foreach ($y as $n => $v) { }

$v (or should I say $x[2]) becomes:

• 'bye bye'
• 'world'
• 'harsh'