PHP Variable Overriding

Question

When I try to Override the class variable same way as override the class method in PHP. Like:

class DataMapper {
     protected $_name = null;

     public function printName() {
          echo $this->_name;
     }
}

class Model extends DataMapper {
     protected $_name = 'Ana';
}

$test = new Model();
$test->printName();

It's print 'Ana'.

Why PHP can do such a thing like that ? It break the law of object oriented paradigm

I don't think the language behaviour is illegal in any jurisdiction. Variables do not actually need to be shadowed. And in PHP objects are quite simply dictionaries too. You only redeclare the default value here. — mario
– mario, Commented Jun 14, 2011 at 4:56
Those are not class variables (i.e. static), they are object/instance variables. As people explain, this is how inheritance is supposed to work with protected members. If you were instead to set a private class variable (i.e. private static $myVar = 'Ana';) then it would not be defined in any child class. — joelhardi
– joelhardi, Commented Jun 14, 2011 at 5:11
The way I understand it, you are right. You normally cannot override variables. And here you are not really doing that, you are simply redeclaring a variable, and the original is lost. I think, anyways. I think PHP is just a little looser in its standards, and lets you get away with questionable syntax/operations. — Jonathon
– Jonathon, Commented Apr 17, 2013 at 1:59

Community · Accepted Answer · 2023-11-17 19:24:28Z

30

It's not. That's how PHP is supposed to work. Have a look at PHP Classes and Objects Visibility.

Objects of the same type will have access to each others private and protected members even though they are not the same instances. This is because the implementation specific details are already known when inside those objects.

Because Model extends DataMapper, it has access to its functions, variables and such but it can override them which is what happened. Although your function lives in the DataMapper class, it's called from (and inherited by) the Model class in which the name is set to Ana.

edited Nov 17, 2023 at 19:24

CommunityBot

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answered Jun 14, 2011 at 4:54

Francois Deschenes

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2 Comments

Hensembryan Over a year ago

wow really nice answer, i wonder if PHP can do this. I've try the relevant code in C++, C#, and JAVA but no luck

MengT Over a year ago

@brian, PHP can do this, which is a language that allows variable overriding; Java only supports variable hiding, so you will get different results.

BraedenP · Accepted Answer · 2011-06-14 05:00:52Z

4

I think you're just having trouble understanding what $this does. When you reference $this, it is actually referencing the current object.

When you inherit the DataMapper class, the printName() method is made accessible inside Model objects, but the $this reference still refers to the current Model object, $test.

Since the $_name property of Model objects is instantiated to "Ana" it is printing Ana. This is exactly what is expected. Perhaps having another read through the theories of Inheritance and Scope would help you out with understanding what's going on here.

answered Jun 14, 2011 at 5:00

BraedenP

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2 Comments

Hensembryan Over a year ago

yeah but in other language like c++, c#, and java it can't do that. I mean override the class variable

drzaus Over a year ago

@brian in c# you can, you just have to be explicit that you intend to override it. otherwise it just hides the parent variable from that scope downwards. the difference in PHP is that the override is implicit.

Balanivash · Accepted Answer · 2011-06-14 04:54:39Z

3

I dont think this breaks the "law of OO". You have inherited the DataMapper class. And thus you have inherited the public function printName(). So when you call the function it acts like the function that belongs to the model class.

answered Jun 14, 2011 at 4:54

Balanivash

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