PHP ArrayAccess - multidimensional array and offsetGet with reference

Question

I read a lot of past questions about ArrayAccess PHP interface and it's method offsetGet that can return a reference. I have a simple class implementing this interface that wraps a variable of type array. The offsetGet method returns a reference, however I get an error saying Only variable references should be returned by reference. Why?

class My_Class implements ArrayAccess {
    private $data = array();

    ...

    public function &offsetGet($offset) {
        return isset( $this->data[ $offset ] ) ? $this->data[ $offset ] : null;
    }

    ...
}

I would like to be able to use multidimensional arrays with this class:

$myclass = new My_Class();

$myclass['test'] = array();
$myclass['test']['test2'] = array();
$myclass['test']['test2'][] = 'my string';

Because you return NULL unbound to a variable, does it help if you rewrite it to if(!isset( $this->data[ $offset ]) $this->data[ $offset ] = null; return $this->data[ $offset ]; ? /too lazy to test — Wrikken
– Wrikken, Commented Nov 24, 2014 at 19:18
Your function returns the result of an expression, not a reference. And a & null reference isn't possible. Try temporary variables. — mario
– mario, Commented Nov 24, 2014 at 19:18

Pierre-Olivier Vares · Accepted Answer · 2017-04-28 12:28:55Z

In this code :

public function &offsetGet($offset) {
    $returnValue = isset( $this->data[ $offset ] ) ? $this->data[ $offset ] : null;
    return $returnValue;
}

$returnValue is a copy of $this->data[$offset], not a reference.

You have to make itself a reference, and for that you have to replace the ternary operator with an if statement :

public function &offsetGet($offset) {
    if (isset($this->data[$offset]) {
        $returnValue &= $this->data[$offset]; // note the &=
    }
    else {
        $returnValue = null;
    }
    return $returnValue;
}

should do the trick.

For the non-existent case, I'd rather throw an Exception, like the one you got when asking for a non-existing key of an array. Since the value you return won't be a reference,

$myclass['non-existing']['test2'] = array();

will presumably throw an indirect overloaded modification error, and should therefore be forbidden.

Arron · Accepted Answer · 2014-11-24 19:18:20Z

0

I think this is becuase you are returning the result of an expression, not a variable. Try writing the if statement out and return the actual variable.

see php manual -> second note

answered Nov 24, 2014 at 19:18

Arron

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Comments

Nalin Singapuri · Accepted Answer · 2014-11-24 19:22:31Z

0

The method '&offsetGet' is returning a reference (pointer) to a variable.

You need to either modify the method signature from '&offsetGet' to 'offsetGet' or use a variable to hold the return value.

// modify method signiture
public function offsetGet($offset) {
    return isset( $this->data[ $offset ] ) ? $this->data[ $offset ] : null;
}

// or use a variable to hold the return value.
public function &offsetGet($offset) {
    $returnValue = isset( $this->data[ $offset ] ) ? $this->data[ $offset ] : null;
    return $returnValue;
}

answered Nov 24, 2014 at 19:22

Nalin Singapuri

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3 Comments

Stefano Over a year ago

I can't use the first statement because it's returning a copy of the array instead of the original. The second statement is throwing a new error: Indirect modification of overloaded element of My_Class has no effect

Nalin Singapuri Over a year ago

Give me the context up the stack. Whats calling offsetGet?

Stefano Over a year ago

It's in the question. I'm creating a new class instance and then trying to set some data in the array

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