PHP Pass by reference in foreach

Question

I have this code:

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

foreach ($a as &$v) {

}

foreach ($a as $v) {
  echo $v.PHP_EOL;
}

Can somebody explain why the output is: zero one two two .

From zend certification study guide.

Answer 1

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'... : )

Answer 2

I'm not sure I can explain why but this is documented in the manual.

The comment also provides some solutions if that is what you are looking for.

Answer 3

Because if you create a reference to a variable, all names for that variable (including the original) BECOME REFERENCES.