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up vote 2 down vote favorite

Why this code not works as I expected? Inside the Test(&$array) function I would set the ref parameter to the global $array1 but this didn't works. 

$array1 = array();
$array2 = array();

function Test(&$array)
{
    global    $array1;
    $array = &$array1;

    $array['inside'] = 'inside';

}

//SET BY THE FUNCTION:

Test($array2); 
$array2['test1'] = 'test1';

var_dump($array1); //array('inside' => 'inside') ** WHERE IS THE 'test1'  key? **
var_dump($array2); //array('test1' => 'test1')   ** WHERE IS THE 'inside' key? **

//SET WITHOUT THE FUNCTION:

$array2 = &$array1;
$array2['test2'] = 'test2';

var_dump($array1); //array('inside' => 'inside', 'test2' => 'test2') ** FINE **
var_dump($array2); //array('inside' => 'inside', 'test2' => 'test2') ** FINE **

EDIT:

It's quite clear that if I changed $array to point to $array1 then $array1 will have the 'inside' => 'inside' value outside the function. What not clear that if I set $array2['test1'] = 'test1' why not change this $array1 also? Its 'linked' before inside the function!

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6  
Ever heard of return values? –  KingCrunch Jan 14 '13 at 20:29
    
@kingcrunch This is clearly a simplified test case, so criticising an unknown use case seems rather unhelpful. –  IMSoP Jan 14 '13 at 20:58
    
@IMSoP OK, sorry :) But to be honest: I've never seen a useful use-case for in-out-params for years now and I fear they simply don't exists (anymore). But I've seen the pass-by-reference for micro-optimizations (beside: it doesn't optimize anything, because of copy-on-write) and then the same developers were confused, because the array "magically" changed, because some others uses array, like one uses array, that were passed as parameter :) What I want to say: No real use-cases, but many "wtf"-momemts are waiting. –  KingCrunch Jan 14 '13 at 21:02
    
@KingCrunch: The real function should return boolean value while set up multiple parameter values by reference. This is a design pattern from my .NET C# solutions. I just like to implement it by php. –  ggabor Jan 14 '13 at 21:07
    
@ggabor This "return values as status parameter" is a bad habit. When your method fails, use exceptions, and when it succeeds, .. well don't throw an exception ;) and simply go on. Thats the reason why Exceptions exists: Inform you (and give you the opportunity to react on) exceptional situations. To be fair: Maybe in other languages it is a more adopted pattern, but at least not in the languages I know :) (I'm not so familar with the C-family) –  KingCrunch Jan 14 '13 at 21:22

2 Answers 2

up vote 1 down vote accepted

Here's my understanding of references in PHP, and why that code doesn't do what you expect (somebody please jump on me if I get it wrong, I frequently do with references! Also, I'm sure there are better terms than what I'm calling "identifier" and "value"; I just wanted to avoid using "variable" for either concept.)

  • There are a bunch of variable identifiers (the way you get at the data), and a bunch of variable values (where the data actually is)
  • For a normal variable, there is just one identifier pointing at one value. e.g. $foo refers to a specific value - a bucket somewhere in PHP's internals that can hold a number, a string, etc
  • Every time you use the normal assignment operator, e.g. $foo = 42, PHP looks for the value being pointed at and updates it - so the identifier $foo hasn't changed, but the value it points at has.
  • When you assign a reference, e.g. $bar =& $foo, you are actually telling PHP to change the identifier itself. So now $bar and $foo are two different identifiers pointing at the same value. $foo = -1 and $bar = -2 will both write to this value, and whichever name you give it, you're referring to that value.
  • So far, so good. But what if I now write $foo =& $bob? Since I'm changing the identifier, not the value, $foo starts pointing at the same value as $bob, but $bar stays where it was. So now, changing $foo won't make any difference to $bar any more.
  • A similar thing happens when you pass in a parameter by reference to a function. So in the example in the question, the line Test($array2) still creates a new identifier inside the function called $array, but it points it at the same value as $array2. However, inside the function is the line $array = &$array1 which takes that new identifier ($array) and points it at the same value as $array1. The old value, which $array2 still points at, hasn't changed.
  • There are other situations that act like this as well. For instance, the global and static keywords create an extra identifier pointing at an existing value. If you write function foo() { global $foo; $bar = 2; $foo =& $bar; }, only the function's local identifier called $foo is updated to point at $bar's value; the global identifier (which happens to also be called $foo, when you're in global scope) still points at its original value.

The reference system for ordinary values in PHP always includes exactly one level of indirection - you cannot create a pointer to a pointer to a pointer in the way that you could in C, for instance.

The only tricky case is objects, which as in many languages have an extra level of indirection all of their own - whereas $foo = 42; $bar = $foo; copies the data representing 42 from $foo's value to $bar's, $foo = new stdClass; $bar = $foo copies an object pointer. So although $foo and $bar still have separate values, and an assignment like $foo = 42 won't have any effect on $bar, $foo->a = 1; and $bar->a = 1 will both end up changing the same object.

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Yep, this sounds right. That's what I was trying to explain :-P –  Rocket Hazmat Jan 16 '13 at 17:26
1  
@RocketHazmat Could probably do with summarising the summary, though - I'm not very good at being succinct... ;) –  IMSoP Jan 16 '13 at 21:35
up vote 4 down vote

When you do $array = &$array1; inside your function, you are changing the value of the local variable $array.

It used to have a reference to $array2, but now it contains a reference to $array1. So when you modify $array, you're modifying $array1.

share|improve this answer
    
It always ties my head in knots working out what exactly =& will do, because you have to distinguish between the variable as a label and the variable as a container for some value. –  IMSoP Jan 14 '13 at 20:43
    
@IMSoP: The & means "reference", so the =& means that the variable holds a "link" to another variable. –  Rocket Hazmat Jan 14 '13 at 20:49
    
So it means that I cannot modify any parameter value outside the function even if it was passed by reference. Every function parameters are read-only even if passed by reference? –  ggabor Jan 14 '13 at 21:04
    
@ggabor: The problem is that you changed where the reference is pointing to. $array used to point to $array2, but you changed it to point to $array1 instead, so that's the array that got modified. If you didn't have the $array = &$array1; line, then $array['inside'] = 'inside'; would've modified $array2. –  Rocket Hazmat Jan 14 '13 at 21:05
1  
@RocketHazmat That is exactly the mistake I (and I'm sure others) make: it is not a reference to another variable, both variables are references to the same value. Use =& again, on either variable, and they will be unlinked. –  IMSoP Jan 14 '13 at 21:09

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