PHP: Passing an array item that not exist by reference adds it as null

Question

we've found a behavior that we don't know if it's right or not.

Please, view the next code:

<?php
$a = [];
$b = [];
$a[] = 'Hello';
$b[] = 'Bye';
if (isset($b[1]))
{
  $b1 = $b[1];
}
$a1 = defaultValue($a[1], 'again');
print_r($a);
print_r($b);
exit;
function defaultValue(&$var, $val)
{
  return isset($var) ? $var : $val;
}

The result was:

Array
(
    [0] => Hello
    [1] =>
)
Array
(
    [0] => Bye
)

Why the item $a[1] was created? Thanks!

Vini · Accepted Answer · 2017-03-22 13:52:04Z

3

You are passing $var into DefaultValue as reference (the & sign). This is place where $a[1] is created and set to null.

http://php.net/manual/en/language.references.php

answered Mar 22, 2017 at 13:52

Vini

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Comments

Gabriel Heming · Accepted Answer · 2017-03-22 13:55:34Z

1

It's explained in the manual: http://php.net/manual/en/language.references.whatdo.php

Note: If you assign, pass, or return an undefined variable by reference, it will get created.

So, when you pass an undefined reference, it'll be created as null. Isset ensures that should be set and different of null. So the value "again" is returned.

answered Mar 22, 2017 at 13:55

Gabriel Heming

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3 Comments

jhernandis Over a year ago

Thanks. I've look for in the manual, obviously not with the right terms.

jhernandis Over a year ago

Thankis @gabriel-heming

Gabriel Heming Over a year ago

@jhernandis You're welcome. That isn't a common knowledge.

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