PHP静态类变量在初始实例化后仍保留值

Question

I have:

class A {
    static $instances = 0;
    public $instance;

    public function __construct() {
        $this->instance = ++self::$instances;
    }
}

$a = new A();
$b = new A();

echo "<pre>";
print_r($a);
print_r($b);
echo "</pre>";

Outputs:

A Object
(
    [instance] => 1
)
A Object
(
    [instance] => 2
)

I know that static variables still retain their values after exiting functions but this is in a class context. How come in this class context the class still retains the variable when a new instance is created.

Any references to official documentation would be helpful.

Answer 1

Static variables are not associated to any particular instance/object of a class. The same is for the static variable inside of the class's method ... (comments in http://php.net/manual/en/language.oop5.static.php )

So if you declare a variable as static inside a function, it's static for the whole class and all of its instances, not for each object.

and

The difference between static and non static members is only that a non static member is tied to an instance of a class although a static member is tied to the class, and not to a particular instance. That is, a static member is shared by all instances of a class although a non static member exists for each instance of class.

Answer 2

I had to read this from the Java documentation to understand why:

When a number of objects are created from the same class blueprint, they each have their own distinct copies of instance variables.

Sometimes, you want to have variables that are common to all objects. This is accomplished with the static modifier. Fields that have the static modifier in their declaration are called static fields or class variables. They are associated with the class, rather than with any object. Every instance of the class shares a class variable, which is in one fixed location in memory. Any object can change the value of a class variable...

Answer 3

新的实例化不会保留静态变量，而是您要增加类的$ instances，然后将结果保存到实例化的$ instance中。