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Are multiple variable assignments done by value or reference?

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天命终不由人
天命终不由人 2020-12-10 00:33 
$a = $b = 0;

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

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  • 执念已碎
    2020-12-10 00:54

    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.

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  • 温柔的废话
    2020-12-10 00:57

    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" }
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  • 一向
    2020-12-10 01:08

    its assigns them both the value of 0

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  • 执笔经年
    2020-12-10 01:11

    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>.

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  • 长发绾君心
    2020-12-10 01:16

    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.

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  • 梦如初夏
    2020-12-10 01:16

    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

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