Adding references in PHP to an array creates an array of references

Question

Not sure if this is considered a bug

$array = ['numbers' => [1, 2, 3]];
foreach ($array as &$numbers) {
    $numbers = [4, 5, 6];
}
var_dump($array);
modify_array($array);
var_dump($array);

function modify_array($array_arg)
{
    $array_arg['numbers'] = [1, 2, 3];
}

Prints

array(1) {
  ["numbers"]=>
  &array(3) {
    [0]=>
    int(4)
    [1]=>
    int(5)
    [2]=>
    int(6)
  }
}
array(1) {
  ["numbers"]=>
  &array(3) {
    [0]=>
    int(1)
    [1]=>
    int(2)
    [2]=>
    int(3)
  }
}

Notice that $array was not passed by reference to modify_array() however the nested array was modified. It sounds logical but not sure if this is documented!

marian0 · Accepted Answer · 2014-09-17 23:47:44Z

4

If you are using foreah with reference you prevent PHP to make a copy of array, so you are working on original array. When you finish foreach loop you should unset() variable used in loop, because it still remains in memory. If you don't do this, you will get unpredictable results like this one presented in your question. Even in documentation you have a warning:

Warning Reference of a $value and the last array element remain even after the foreach loop. It is recommended to destroy it by unset().

Here you can find really good examples of foreach loops without unsetting.

So if you add unset(), everything looks as it should

$array = ['numbers' => [1, 2, 3]];
foreach ($array as &$numbers) {
    $numbers = [4, 5, 6];
}
unset($numbers);
var_dump($array);
modify_array($array);
var_dump($array);

function modify_array($array_arg) {
    $array_arg['numbers'] = [1, 2, 3];
}

Result:

array(1) { ["numbers"]=> array(3) { [0]=> int(4) [1]=> int(5) [2]=> int(6) } } 
array(1) { ["numbers"]=> array(3) { [0]=> int(4) [1]=> int(5) [2]=> int(6) } }

answered Sep 17, 2014 at 23:47

marian0

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3 Comments

Jonathan Kuhn Over a year ago

Just want to say that if anyone wants confirmation: codepad.viper-7.com/hcpoWB Only thing I added was the line with br tag.

zerkms Over a year ago

"If you don't do this, you will get unpredictable results" --- what tells us the results must be unpredictable? The warning you quoted does not explain the behaviour.

StackSlave Over a year ago

I agree, the results would be very predictable according to the Warning, but this is still a useful answer besides that.

apoq · Accepted Answer · 2014-09-17 23:44:08Z

0

If you rewrite following code:

foreach ($array as &$numbers) {
    $numbers = [4, 5, 6];
}

To smth like this:

  $array['numbers'] = &$array['numbers'];
  $array['numbers'] = [4,5,6];

It'll be easier to understand such a behavior, and this seems to be a known issue. (See: http://php.net/manual/en/language.references.php#87532)

answered Sep 17, 2014 at 23:44

apoq

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