Are multiple variable assignments done by value or reference?
$a = $b = 0;
In the above code, are both $a and $b assigned the value of 0, or is $a just referencing 开发者_Go百科$b?
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" }
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.
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.
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.
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>.
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
its assigns them both the value of 0
加载中,请稍侯......
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