PHP arrays work differently for objects vs values [duplicate]

Question

Here's my code:

$words = array();
$word = "this";
$words[] = $word;
$word = "that";
$words[] = $word;
print_r($words);

class word{
    private $text;
    public function __construct($word){
        $this->text=$word;
    }

    public function setWord($word){
        $this->text=$word;
    }
}

$class_words = array();
$word = new word("this");
$class_words[] = $word;
$word->setWord("that");
$class_words[] = $word;
print_r($class_words);
exit;

Here's the output:

Array
(
    [0] => this
    [1] => that
)
Array
(
    [0] => word Object
        (
            [text:word:private] => that
        )

    [1] => word Object
        (
            [text:word:private] => that
        )

)

I expected the second output to match the first in that the array should store 'this' and 'that'. It seems array_name[] = <item> makes a copy to the item when it's an array of values but not so when it's an array of objects. How do I make it copy the object to the array instead copying a reference to the object? Do I need to create a new object each time I need to add an object to the array?

That's because that's how String Arrays and Objects work. They are difficult to mxi — Forbs
– Forbs, Commented Mar 18, 2019 at 21:48

Nick · Accepted Answer · 2019-03-18 21:53:33Z

1

If you want to copy the value of the object into the array you need to write a "getter" for the value e.g.

class word{
    private $text;
    public function __construct($word){
        $this->text=$word;
    }

    public function setWord($word){
        $this->text=$word;
    }

    public function getWord() {
        return $this->text;
    }
}

$class_words = array();
$word = new word("this");
$class_words[] = $word->getWord();
$word->setWord("that");
$class_words[] = $word->getWord();
print_r($class_words);

Output:

Array
(
    [0] => this
    [1] => that
)

Demo on 3v4l.org

answered Mar 18, 2019 at 21:53

Nick

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Comments

wowest · Accepted Answer · 2019-03-18 22:17:42Z

1

$x = new X(); stores a reference to an object into $x. A subsequent $y = $x; copies the reference, not the object, so $x and $y both refer to the same object.

PHP has rather complex semantics regarding references.

answered Mar 18, 2019 at 22:17

wowest

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Comments

Dinu · Accepted Answer · 2019-03-18 21:50:02Z

0

Objects are always references; all your uses of $word refer to the same object and the same data structure. You need to do:

$class_words=[new word('this'),new word('that')];

answered Mar 18, 2019 at 21:50

Dinu

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Comments

Tim Morton · Accepted Answer · 2019-03-19 01:00:02Z

0

Both elements of your array hold the same object. So whenever you make a change in that object, all references to that object will show that change-- they're all referring to the same object in its current state.

As mentioned previously, if you want different values, you need to either instantiate new objects, or get the property value via a getter() that returns a simple value (string, integer, boolean, etc), not the object itself.

As a side note, you can leverage the referential nature of objects to chain methods ($Obj->method1()->method2()->method3()) by having a method return a reference to the object, i.e., return $this;

answered Mar 19, 2019 at 1:00

Tim Morton

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