[SOLVED] Return null by reference via _

Return null by reference via __get()

Quick specs:

PHP 5.3
error_reporting(-1) // the highest

I'm using the __get() by reference trick to magically access arbitrarily deep array elements in an object.

Quick example:

public function &__get($key){
    return isset($this->_data[$key])
        ? $this->_data[$key]
        : null;
}

This doesn't work as when the $key isn't set, it tries to return null by reference, which of course throws Only variable references should be returned by reference ... I tried modifying it as follows:

public function &__get($key){
    $null = null;
    return isset($this->_data[$key])
        ? $this->_data[$key]
        : $null;
}

Still doesn't work though, I'm assuming that setting $null to null essentially unset()s it.

What can I do? Thanks!

Just figured I'd promote this question, as it's somewhat relevant (PHP magic and references); __callStatic(), call_user_func_array(), references, and PHP 5.3.1. I've yet to find an answer ...besides modifying the PHP core.

Solution

This has nothing to do with null, but rather the ternary operator:

Rewriting it with an if/else won't throw the notice:

public function &__get($key)
{
    $null = null;
    if (isset($this->_data[$key])) {
        return $this->_data[$key];
    } else {
        return $null;
    }
}

Ternary operators cannot result in references. They can only return values.