Returning elements with php functions

Question

I have a class like this:

class Foo {
    $elements = array();

    function getElementByName($name) {
        foreach($this->elements as $elm) {
            if ($elm->name == $name) {
                return $elm;
            }
        }
    }
}

I expected the following code to modify the element of my array:

$myFoo = new Foo();
$myFoo->getElementByName('foo1')->active = true;

Instead, when running my code, the active property of $elements['foo1'] is still false as it was before calling getElementByName

I think that the function makes a "copy" of the element, how can I get the real element of the array, so that when I modify it, and then access it in the array, its values have changed?

You assume right. But note that PHP since version 5 should pass objects by reference anyway. Are you still running PHP 4? — knittl
– knittl, Commented Oct 13, 2013 at 10:58
@knittl PHP 4.4.9 will throw unexpected T_OBJECT_OPERATOR. — ComFreek
– ComFreek, Commented Oct 13, 2013 at 11:01

kero · Accepted Answer · 2013-10-13 11:00:26Z

3

Return a reference to it (notice the &):

function &getElementByName($name) { ... }

edited Oct 13, 2013 at 11:00

kero

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answered Oct 13, 2013 at 10:55

OneOfOne

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Comments

knittl · Accepted Answer · 2013-10-13 10:57:23Z

1

Return a reference to the element:

function &getElementByName($name) { // & returns by reference
    foreach($this->elements as $elm) {
        if ($elm->name == $name) {
            return $elm;
        }
    }
}

Starting with PHP 5 objects are passed by reference by default.

answered Oct 13, 2013 at 10:57

knittl

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1 Comment

Mark Over a year ago

Actually, object references are passed by value.

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