It seems that I am being able to create a property for a class directly from the instance in PHP?

Question

In the below program the last statement echo $objb->test; should not output any value. The second last statement should throw an error saying that private member cannot be accessed. However, I am getting output 20 when running the program.

Since PHP is loosely type language is the $objb create a new variable outside the class as well...

<?php
    error_reporting(E_ALL);
    class A {
        public $a;
        private $test;

        protected function sayhello(){
            echo "<p>hello class A</p>";
        }

    }//end of class A

    class B extends A {
        //private $c;
        public function sayhellonew(){
            $this->sayhello();
            echo "<p>hello class B</p>";
        }
    }//end of class B

    $objb = new B();
    $objb->sayhellonew();

    $objb->a = 10;
    echo $objb->a;

    echo "<br>";

    $objb->test = 20;
    echo $objb->test;
    ?>

Output I am getting output 20 why is it working?

Answer 1

Since the $test property is private to A, it is not visible in the scope of B. That means that B can have its own property named $test. When you assign objb->test = 20, that's what you're setting.

The following shows that you're not actually setting the private variable:

class A {
    public $a;
    private $test = 10;

    protected function sayhello(){
        echo "<p>hello class A</p>";
    }
    public function showTestA() {
        echo "Test in A = " . $this->test . '<br>';
    }

}//end of class A

class B extends A {
    //private $c;
    public function sayhellonew(){
        $this->sayhello();
        echo "<p>hello class B</p>";
    }
    public function showTestB() {
        echo "Test in B = " . $this->test . '<br>';
    }
}//end of class B

$objb = new B();
$objb->test = 20;
$objb->showTestA();
$objb->showTestB();

This will display:

Test in A = 10
Test in B = 20

Answer 2

Because, you are overriding the parent class' private property with childs class' public property.

It is valid in PHP.

Try adding a new private property $c in class B and try to override it.

You will get an error: Fatal error: Cannot access private property B::$c

Check it here: http://codepad.org/74IgLqYn

Answer 3

Because its not the test same variable that you have declare in it's parent(A) class. In fact its will treat it as public $test; of class B. Check below you will get the issue.

<?php
    error_reporting(E_ALL);
    class A {
        public $a;
        private $test;

        protected function sayhello(){
            echo "<p>hello class A</p>";
        }

        public function setTest(){
            $this->test = "50";
        }

    }//end of class A

    class B extends A {
        //private $c;
        public function sayhellonew(){
            $this->sayhello();
            echo "<p>hello class B</p>";
        }
    }//end of class B

    $objb = new B();
    $objb->sayhellonew();

    $objb->a = 10;
    echo $objb->a;

    echo "<br>";
    $objb->setTest();
    echo $objb->test;
    ?>

CODEPAD.