Are multiple variable assignments done by value or reference?

问题

$a = $b = 0;

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

回答1:

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" }

回答2:

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.

回答3:

You could have tried it yourself

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

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

(currently I dont really care :D)

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

回答4:

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.

回答5:

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

回答6:

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>.

回答7:

its assigns them both the value of 0

来源：https://stackoverflow.com/questions/6257131/are-multiple-variable-assignments-done-by-value-or-reference

标签

php

variables

variable-assignment