Are multiple variable assignments done by value or reference?

Question

$a = $b = 0;

In the above code, are both $a and $b assigned the value of 0, or is $a just referencing $b?

Bob_Gneu · Accepted Answer · 2012-01-25T00:50:57.813

57

With raw types this is a copy.

test.php

$a = $b = 0;

$b = 3; 

var_dump($a);
var_dump($b);

Output:

int(0) 
int(3)

With objects though, that is another story (PHP 5)

test.php

class Obj
{ 
    public $_name;
}

$a = $b = new Obj();

$b->_name = 'steve';

var_dump($a);
var_dump($b);

Output

object(Obj)#1 (1) { ["_name"]=> string(5) "steve" } 
object(Obj)#1 (1) { ["_name"]=> string(5) "steve" }

edited Jan 25 '12 at 00:50

answered Jun 06 '11 at 19:47

Bob_Gneu

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9

Note, that the example is misleading: Changing the property of an object is not the same as changing the variable the object is referenced from. Because the variable never changed, it is _of course_ the same. – KingCrunch Dec 30 '11 at 19:00
1

According to [documentation](http://php.net/manual/en/internals2.variables.intro.php), PHP uses a copy on write system. It means that with primitive types such as strings or ints are really created in memory on change. That's why, contrary to what we might think, creating references to optimize performance on small variables may not bring the desired effect, because PHP needs to create it before. However, objects are passed by reference by default since PHP5. – bgondy Apr 11 '13 at 17:00
4

The assign by reference with objects can be changed to assign by copy using `clone`. EG: `$a = clone $b = new Obj;` – Will B. Mar 31 '16 at 21:17
At the class variable declaration level how it would be ? Like this `private $_name = $address = "";` – Chaminda Bandara Nov 14 '16 at 10:12

score 20 · Answer 2 · answered Jun 06 '11 at 19:46

20

Regard this code as:

$a = ($b = 0);

The expression $b = 0 not only assigns 0 to $b, but it yields a result as well. That result is the right part of the assignment, or simply the value that $b got assigned to.

So, $a gets assigned 0 as well.

answered Jun 06 '11 at 19:46

Blagovest Buyukliev

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score 9 · Answer 3 · answered Jun 06 '11 at 19:45

9

You could have tried it yourself

$a = $b = 0;
$a = 5;
echo $b;

or

$a = $b = 0;
$b = 5;
echo $a;

(currently I dont really care :D)

Thus: No, they are both independent variables with the value 0.

answered Jun 06 '11 at 19:45

KingCrunch

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In that example, wouldn't you have to assign 5 to $b on the second line to prove the reference was not present? and yes, you are correct. I figured having the question up here would be nice for others searching as well though... – Evil Elf Jun 06 '11 at 19:47
@Evil: Don't want to think about it, so added the other way round too. Both return `0`, thus it _must_ be true in any case :) – KingCrunch Jun 06 '11 at 19:49

Tomasz Kowalczyk · Answer 4 · 2011-06-06T20:25:52.497

I'll recommend a good read on this: http://terriswallow.com/weblog/2007/multiple-and-dynamic-variable-assignment-in-php/ . In one of comments, you can read:

It should be noted that if you use multiple assignment on one line to assign an object, the object is assigned by reference. Therefore, if you change the value of the object’s property using either variable, the value essentially changes in both.

So I'll personally recommend that you assign the variables separately.

For the record:

$a = $b = 4;
var_dump($a, $b);
$b = 5;
var_dump($a, $b);

Yields:

int(4)
int(4)
int(4)
int(5)

But:

class Tmp
    {
    public $foo;

    public function __construct()
        {
        $this->foo = 'bar';
        }
    }

$a = $b = new Tmp();
var_dump($a, $b);
$a->foo = 'oth';
var_dump($a, $b);

Yields:

object(Tmp)#1 (1) {
  ["foo"]=>
  string(3) "bar"
}
object(Tmp)#1 (1) {
  ["foo"]=>
  string(3) "bar"
}
object(Tmp)#1 (1) {
  ["foo"]=>
  string(3) "oth"
}
object(Tmp)#1 (1) {
  ["foo"]=>
  string(3) "oth"
}

So the conclusion is that there is no reference for primitives, but there IS a reference to objects.

we proved this false actually with some tests. – Evil Elf Jun 06 '11 at 20:15 — Evil Elf, Jun 06 '11 at 20:15

score 4 · Answer 5 · answered Feb 21 '15 at 18:05

It depends what're you assigning.

If you're assigning a value, then the assignment copies the original variable to the new one.

_{Example 1:}

$a = $b = 0;
$b++; echo $a;

Above code will return 0 as it's assignment by value.

_{Example 2:}

$a = ($b = 4) + 5; // $a is equal to 9 now, and $b has been set to 4.

An exception to the usual assignment by value behaviour within PHP occurs with objects, which are assigned by reference in PHP 5 automatically. Objects may be explicitly copied via the clone keyword.

_{Example 3}

$a = $b = $c = new DOMdocument();
$c->appendChild($c->createElement('html'));
echo $a->saveHTML();

Above code will print <html></html>.

score 2 · Answer 6 · answered Jun 06 '11 at 19:48

2

Both $a and $b are assigned that value of 0. If you wanted $a to reference $b, you would preempt it with an ampersand, e.g.:

$a = & $b = 0;

http://php.net/manual/en/language.oop5.basic.php

answered Jun 06 '11 at 19:48

Taylor Gerring

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score 1 · Answer 7 · answered Jun 06 '11 at 19:46

1

its assigns them both the value of 0

answered Jun 06 '11 at 19:46

mcgrailm

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Are multiple variable assignments done by value or reference?

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