PHP Inheritance Accessing Variables

Question

So I have this code, see below. I am running an error while getting a protected value. Thought that protected values will be inherited. I was figuring out why we need to call parent::__construct(), while we just can extend a class.

If you could tell me whats wrong and how can I do it the right way, that would be awesome.

    <?php 
    /**
    * DogWords
    */

    class DogWords
    {
        protected $words = array('Wrr', 'Grr', 'Weeee', 'Houou');
    }

    /**
    * Dog Class
    */
    class Dog extends DogWords
    {
        protected $dogname = NULL;
        protected $dogwords = NULL;

        public function __construct($dogname)
        {
            $this->dogname = $dogname;
            $this->dogwords = new DogWords;
        }

        public function bark()
        {
            echo $this->dogname . "; Bark, bark, bark...";
        }
    }

    /**
    * Poodle
    */
    class Poodle extends Dog
    {

    }

    $Amy = new Poodle('DogConstructor');
    echo $Amy->dogwords->words[1];      // Fatal Error...
    echo $Amy->bark();      // DogConstructor; Bark, bark, bark...
 ?>

You should not have Dog extend DogWords. That's just silly. — cwallenpoole
– cwallenpoole, Commented Oct 21, 2013 at 17:50
Sure, Dog should inherit from Animals, Mammals, or HomePets. Because it should inherit from a class that is more abstract than dog itself. - For further readers... — user2904228
– user2904228, Commented Oct 21, 2013 at 19:23
Yes, that is correct. The problem is that you have an object inheriting from <name>Words. Having an abstract class which has a words property is one thing, having a dog be a subclass of bark is just wrong. — cwallenpoole
– cwallenpoole, Commented Oct 21, 2013 at 20:56

Arun Killu · Accepted Answer · 2013-10-21 17:49:02Z

2

Protected members cannot be used outside the scope of the subclass , if you want to access it outside it should be declared as public .

Please study from here docs

answered Oct 21, 2013 at 17:49

Arun Killu

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3 Comments

user2904228 Over a year ago

Ohh you right. I've seen in some frameworks that they use B extends A and then they use again parent::__construct(). But why would you do that exactly?

Arun Killu Over a year ago

tats different that's for calling the parent constructor from derived class

user2904228 Over a year ago

But why would you call the constructor of the parent class when extending that class at the same time? When you make instance of B even when parent::__construct() will not be specified in B, PHP will still call the constructor of A because that's the nearest one. And B will still have all the properties and methods 'A' has.

Alex · Accepted Answer · 2013-10-21 17:56:36Z

If you pretend to keep the variable to protected, why just don't do the following:

class DogWords
{
    protected $words = array('Wrr', 'Grr', 'Weeee', 'Houou');
}

/**
* Dog Class
*/
class Dog extends DogWords
{
    protected $dogname;
    protected $dogwords;

    public function __construct($dogname)
    {
        $this->dogname = $dogname;
        $this->dogwords = new DogWords;
    }

    public function bark()
    {
        return $this->dogname . "; Bark, bark, bark...";
    }
}

/**
* Poodle
*/
class Poodle extends Dog
{
     public function getDogWords()
     {
         return $this->dogwords->words;
     }
}

$Amy = new Poodle('DogConstructor');
$words = $Amy->getDogWords();
var_dump($words[1]);
var_dump($Amy->bark());

Example: http://codepad.org/GsmmHxev

user4035 · Accepted Answer · 2013-10-21 17:52:44Z

0

The error comes from this fact: "Members declared protected can be accessed only within the class itself and by inherited and parent classes".

http://www.php.net/manual/en/language.oop5.visibility.php

In order to access them, you need to make them both public:

class DogWords
{
    public $words = array('Wrr', 'Grr', 'Weeee', 'Houou');
}

/**
 * Dog Class
 */
class Dog extends DogWords
{
    protected $dogname = NULL;
    public $dogwords = NULL;

//... rest of the code

$Amy = new Poodle('DogConstructor');
echo $Amy->dogwords->words[1];      // Fatal Error...
echo $Amy->bark();      // DogConstructor; Bark, bark, bark...
?>

Now this prints: "GrrDogConstructor; Bark, bark, bark..."

answered Oct 21, 2013 at 17:52

user4035

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Comments

Rajesh Paul · Accepted Answer · 2013-10-21 18:55:57Z

0

According to the rule of OOP, a protected member can only be accessed from inside the base class and the derived class(immediate and non-immediate) but not from anywhere else.

But here you are trying to access the protected member of class dog i.e. dogwords in echo $Amy->dogwords->words[1] using the subclass poodle's object i.e. Amy. Hence according to the principle you cannot access it this way.

Perfectly legal...

answered Oct 21, 2013 at 18:55

Rajesh Paul

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