Notice: Only variable references should be returned by reference

Question

I'm getting this notice:

Notice: Only variable references should be returned by reference

private function &getPageByLabel(&$pages, $label){
    foreach($pages as &$page){
        if($page['label'] == $label) return $page;
    }
    return false; //THIS IS THE LINE THAT THE ERROR MESSAGE MENTIONS
}

I'm using ZF 1.12 and PHP 5.5.9

`&$page` is wrong because the index can't be used as acumulator, should be `$pages` instead — Fernando Torres, Sep 09 '19 at 17:01
I have never seen a function reference like this. `&getPageByLabel` looks wrong to me — aynber, Sep 09 '19 at 17:01
@aynber That it's true, `&getPageByLabel` is wrong and need to be removed the & symbol — Fernando Torres, Sep 09 '19 at 17:02
Please check my answer as I've updated it since first posting with more details. — Jochem Kuijpers, Sep 09 '19 at 17:13
Possible duplicate of [PHP Multidimensional Array Searching (Find key by specific value)](https://stackoverflow.com/questions/8102221/php-multidimensional-array-searching-find-key-by-specific-value) — Dharman, Sep 09 '19 at 17:19

Jochem Kuijpers · Answer 1 · 2019-09-09T17:12:51.267

Your function specifies that it returns a reference to a variable (meaning it does not copy the value, it passes exactly the version that you have inside the function).

Since you return a constant, there is no variable associated with this, so it should not work. PHP is kind though, and interprets what you meant to write as:

private function &getPageByLabel(&$pages, $label){
    foreach($pages as &$page){
        if($page['label'] == $label) return $page;
    }
    $return_value = false;
    return $return_value;
}

Removing the & before getPageByLabel will remove the error, since now you will be returning a (copied) value instead of a reference.

Though note that this may cause your code to malfunction, since you were expecting a reference to one of your $pages, likely to modify it externally. Without the reference, you will just modify the copy externally and the $page in $pages will remain unchanged.

A better way is probably to create a 'dummy object' to return.

You could make a 'dummy page' named $no_page and return that instead of false. Then externally you should be able to determine whether or not the page exists by:

$page = $x->getPageByLabel(...); 
if ($page === $x->no_page) {
    // no page found!
}

A cleaner option is to check for existence first though it is slightly slower:

private function pageExistsWithlabel(&$pages, $label) {
    foreach ($pages as $page) {
        if ($page['label'] == $label) return true;
    }
    return false;
}

private function &getPageByLabel(&$pages, $label){
    foreach($pages as &$page){
        if($page['label'] == $label) return $page;
    }
    // throw an error here
}

if (! $x->pageExistsWithlabel(...)) {
    // handle the case where the page does not exist to prevent the error
} else {
    $page = $x->getPageByLabel(...);
}

Notice: Only variable references should be returned by reference

1 Answers1