PHP Pass by reference in foreach

Because on the second loop, $v is still a reference to the last array item, so it's overwritten each time.

You can see it like that:

$a = array ('zero','one','two', 'three');

foreach ($a as &$v) {

}

foreach ($a as $v) {
  echo $v.'-'.$a[3].PHP_EOL;
}

As you can see, the last array item takes the current loop value: 'zero', 'one', 'two', and then it's just 'two'... : )

I had to spend a few hours to figure out why a[3] is changing on each iteration. This is the explanation at which I arrived.

There are two types of variables in PHP: normal variables and reference variables. If we assign a reference of a variable to another variable, the variable becomes a reference variable.

for example in

$a = array('zero', 'one', 'two', 'three');

if we do

$v = &$a[0]

the 0th element ($a[0]) becomes a reference variable. $v points towards that variable; therefore, if we make any change to $v, it will be reflected in $a[0] and vice versa.

now if we do

$v = &$a[1]

$a[1] will become a reference variable and $a[0] will become a normal variable (Since no one else is pointing to $a[0] it is converted to a normal variable. PHP is smart enough to make it a normal variable when no one else is pointing towards it)

This is what happens in the first loop

foreach ($a as &$v) {

}

After the last iteration $a[3] is a reference variable.

Since $v is pointing to $a[3] any change to $v results in a change to $a[3]

in the second loop,

foreach ($a as $v) {
  echo $v.'-'.$a[3].PHP_EOL;
}

in each iteration as $v changes, $a[3] changes. (because $v still points to $a[3]). This is the reason why $a[3] changes on each iteration.

In the iteration before the last iteration, $v is assigned the value 'two'. Since $v points to $a[3], $a[3] now gets the value 'two'. Keep this in mind.

In the last iteration, $v (which points to $a[3]) now has the value of 'two', because $a[3] was set to two in the previous iteration. two is printed. This explains why 'two' is repeated when $v is printed in the last iteration.

First loop

$v = $a[0];
$v = $a[1];
$v = $a[2];
$v = $a[3];

Yes! Current $v = $a[3] position.

Second loop

$a[3] = $v = $a[0], echo $v; // same as $a[3] and $a[0] == 'zero'
$a[3] = $v = $a[1], echo $v; // same as $a[3] and $a[1] == 'one'
$a[3] = $v = $a[2], echo $v; // same as $a[3] and $a[2] == 'two'
$a[3] = $v = $a[3], echo $v; // same as $a[3] and $a[3] == 'two'

because $a[3] is assigned by before processing.