PHP Object Question

Unfortunately I cannot provide any code examples, however I will try and create an example. My question is about Objects and memory allocation in PHP.

If I have an object, lets say:

$o开发者_运维问答bject = new Class();

Then I do something like

$object2 = $object;

What is this actualy doing? I know there is a clone function, but thats not what I'm asking about, I'm concerned about whether this is creating another identical object, or if its just assigning a reference to $object.

I strongly understand this to mean that it just creates a reference, but in some case usages of mine, I find that I get another $object created, and I can't understand why.

If you use the magic method __invoke, you can call an object similar to a function, and it will call that magic method.

class Object{
    function __invoke(){ return "hi"; }
}

$object = new Object;
$object2 = $object();
echo $object2; // echos hi

That means that $object2 is equal to whatever that function returns.

Basically, you are calling a function, but using a variable as it's name. So:

function test(){ echo "hi"; }
$function_name = "test";
$function_name(); // echos hi.

In this case, you are just calling an object instead.

So, in reference to your question, this is actually not 'cloning' at all, unless the __invoke() function looks like this:

function __invoke(){ return this }

In which case, it would be a reference to the same class.

You are creating a second reference of the same object. Here is a proof:

<?php
class TestClass {
    private $number;

    function __construct($num) { $this->number = $num; }
    function increment() { $this->number++; }

    function __toString() { return (string) $this->number; }
}

$original = new TestClass(10);

echo "Testing =\n";
echo "--------------------------------\n";
echo '$equal = $original;' . "\n";
$equal = $original;

echo '$equal = ' . $equal . ";\n";

echo '$original->increment();' . "\n";
$original->increment();

echo '$equal = ' . $equal . ";\n";

echo "\n";
echo "Testing clone\n";
echo "--------------------------------\n";
echo '$clone = clone $original;' . "\n";
$clone = clone $original;

echo '$clone = ' . $clone . ";\n";

echo '$original->increment();' . "\n";
$original->increment();

echo '$clone = ' . $clone . ";\n";

Use clone if you want to create a copy of an instance.

Assuming that you mean

$object2 = $object;

And not

$object2 = $object(); 

PHP will create a reference to the original object, it will not copy it. See http://www.php.net/manual/en/language.oop5.basic.php, the section called Object Assignment.

<?php

class Object{
    public $value = 1;

    public function inc(){
        $this->value++;
    }
}

$object = new Object;
$object2 = $object;

$object->inc();

echo $object2->value; // echos 2, proving it's by reference