Using undefined property in php class

Question

<?php
class a{
    public function out(){
        $this->test = 8;
        return $this->test;
    }
}
$b = new a();
echo $b->out();
?>

output: 8

when i run this code, output the result 8 .

but when i add __set() function, it output a notice, and not 8 output

<?php
class a{
    public function __set($property, $value) {  
    }
    public function out(){
        $this->test = 8;
        return $this->test;
    }
}
$b = new a();
echo $b->out();
?>

output:

PHP Notice: Undefined property: a::$test in /usercode/file.php on line 13

why is it happening?

First define $test variable in constructor and then assign $this->test="8"; hope it will work — Soniya Basireddy
– Soniya Basireddy, Commented Jan 25, 2018 at 9:29
Because that's what __set does. By default, PHP will create properties on the fly. If you've defined __set and it doesn't then set the property within the method, then the property won't exist, and will raise a notice when you try and read it. — iainn
– iainn, Commented Jan 25, 2018 at 9:30

Kevin Kopf · Accepted Answer · 2018-01-25 11:20:01Z

10

As per the docs

__set() is run when writing data to inaccessible properties.

Since you do not have anything in your __set body, the property is not created and therefore not available. You have to define the method body.

class a{
    public function __set($property, $value) {
        $this->$property = $value;
    }
    public function out(){
        $this->test = 8;
        return $this->test;
    }
}
$b = new a();
echo $b->out();

Now THAT outputs 8.

Update You are asking why the first block of code works and the second does not. Take a look at PHP source code here and you will see the explanation in the code itself.

Looks to me, that when you do not have __set() in your class and you do $this->test, PHP internally calls it's own __set(), which does exactly what it does: sets the property name to certain value.

But when you define __set() with empty body, it overrides the default internal __set() and does nothing. And that is the main reason for your code to fail - the requested property has not been set neither by your __set(), nor by the internal one.

edited Jan 25, 2018 at 11:20

answered Jan 25, 2018 at 9:34

Kevin Kopf

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2 Comments

Deqin Chen Over a year ago

Thank you for your answer. I understand what you said. But I want to know why my first program can run successfully. Is there any official explanation?

Kevin Kopf Over a year ago

@DeqinChen I actually updated my answer, linking to PHP source code as well, so that you could see the reason for yourself.

axiac · Accepted Answer · 2018-01-25 10:01:21Z

6

When a::out() runs, there is no $test property in the object. This is why $this->test = 8; invokes a::__set().

But a::__set() doesn't create the $test property and the next statement (return $this->test;) cannot find it and produces the notice.

You should declare the object properties in the class definition and initialize them in the constructor (if appropriate):

class a {
    private $test;               // <-- because of this, $this->test exists...

    public function __set($property, $value) {  
    }

    public function out() {
        $this->test = 8;         // ... and __set() is not invoked here
        return $this->test;
    }
}

Without __set() being defined, the statement $this->test = 8; creates the $test property of the current object if it is not already created (by its definition or by a previous assignment to it) then stores 8 into it.

When __set() is defined, any attempt to set a property that doesn't exist or it is not accessible (setting inside the class a private property inherited from the parent class or setting a protected or private property outside the class) is handled by __set(). Your implementation of __set() doesn't create the missing property and it basically turns the statement $this->test = 8; into a no-op.

edited Jan 25, 2018 at 10:01

answered Jan 25, 2018 at 9:34

axiac

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3 Comments

Deqin Chen Over a year ago

Thank you for your answer. I understand what you said. But I want to know why my first program can run successfully. Is there any official explanation?

axiac Over a year ago

@DeqinChen I updated the answer. Also read the documentation of __set().

Raph Over a year ago

__set IS invoked but it does not do anything, therefore, nothing is set. Thus the warning on the return.

winlans · Accepted Answer · 2018-01-25 09:41:18Z

2

The following is true.

<?php
class a{
    public function __set($property, $value) {
        $this->$property = $value;
    }
    public function out(){
        $this->test = 8;
        return $this->test;
    } 
}
$b = new a();
echo $b->out();

you should look at is php overloading

Find the answers in the manual.

answered Jan 25, 2018 at 9:41

winlans

915 bronze badges

1 Comment

Deqin Chen Over a year ago

Thank you for your answer. I understand what you said. But I want to know why my first program can run successfully. Is there any official explanation?

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