Passing/Returning references to object + changing object is not working

Question

I am using an answer from How to get random value out of an array to write a function that returns a random item from the array. I modified it to pass by reference and return a reference.

Unfortunately, it doesn't appear to work. Any modifications to the returned object do not persist. What am I doing wrong?

I'm on PHP 5.4 if that makes a difference (Don't ask).

function &random_value(&$array, $default=null)
{
    $k = mt_rand(0, count($array) - 1);
    $return = isset($array[$k])? $array[$k]: $default;
    return $return;
}

Usage...

$companies = array();
$companies[] = array("name" => "Acme Co",  "employees"=> array( "John", "Jane" ));
$companies[] = array("name" => "Inotech",  "employees"=> array( "Bill", "Michael" ));

$x = &random_value($companies);
$x["employees"][] = "Donald";
var_dump($companies);

Output...

array(2) {
  [0] =>
  array(2) {
    'name' =>
    string(7) "Acme Co"
    'employees' =>
    array(2) {
      [0] =>
      string(4) "John"
      [1] =>
      string(4) "Jane"
    }
  }
  [1] =>
  array(2) {
    'name' =>
    string(7) "Inotech"
    'employees' =>
    array(2) {
      [0] =>
      string(4) "Bill"
      [1] =>
      string(7) "Michael"
    }
  }
}

I even copied and pasted the examples from the documentation and none of those worked either. They all output null.

Client requirement! I can only tell them so many times in so many ways that being on 5.4 is a terrible idea. — Jeff
– Jeff, Commented Sep 21, 2015 at 18:57
Why are you using references? Are you just trying to return a random array from $companies? — AbraCadaver
– AbraCadaver, Commented Sep 21, 2015 at 19:04
@bishop Needs more red! I actually showed them that and it hasn't changed their mind yet. — Jeff
– Jeff, Commented Sep 21, 2015 at 19:13

bishop · Accepted Answer · 2015-11-10 19:54:18Z

3

The ternary operator creates an implicit copy, which breaks the reference chain. Use an explicit if... else:

function &random_value(&$array, $default=null)
{
    $k = mt_rand(0, count($array) - 1);
    if (isset($array[$k])) {
        return $array[$k];
    } else {
        return $default;
    }
}

As to why, the docs now state:

Note: Please note that the ternary operator is an expression, and that it doesn't evaluate to a variable, but to the result of an expression. This is important to know if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a return-by-reference function will therefore not work and a warning is issued in later PHP versions.

See also this bug where the ternary operator actually returns by reference in a foreach context, when it shouldn't.

edited Nov 10, 2015 at 19:54

answered Sep 21, 2015 at 19:00

bishop

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6 Comments

Jaime Over a year ago

I just tried that. I returns the same incorrect result.

Jeff Over a year ago

It works for me too! I will have to read up on why the ternary operator does that. I wouldn't expect that to happen (obviously!). Thanks!

Jaime Over a year ago

Hmm. Your function still returns the whole array. He only wants one element doesn't he?

bishop Over a year ago

It's how the zval copy-on-write processing happens in the engine, at least up through 5.6. See also this post. I'm not sure what if anything changed in 7. (I doubt it.)

bishop Over a year ago

@Jaime: returns only one element for me and 3v4l. Note that each element is an array of two keys: name and employees.

|

AbraCadaver · Accepted Answer · 2015-09-21 19:26:02Z

3

I'm having trouble wrapping my head around a better function aspect than @bishop as I have just consumed an enormous lunch, however this works:

$x =& $companies[array_rand($companies)];
$x["employees"][] = "Donald";
var_dump($companies);

answered Sep 21, 2015 at 19:26

AbraCadaver

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1 Comment

bishop Over a year ago

I would recommend this approach also, but I suppose the mitigating factor is the use of $default for null values, eg: $companies = [ [], ['name' => 'Foo', 'employees' => []] ];

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